五题中AC了两题，排名1757/5132。

Pig is visiting a friend.

Pig’s house is located at point 0, and his friend’s house is located at point m on an axis.

Pig can use teleports to move along the axis.

To use a teleport, Pig should come to a certain point (where the teleport is located) and choose where to move: for each teleport there is the rightmost point it can move Pig to, this point is known as the limit of the teleport.

Formally, a teleport located at point x with limit y can move Pig from point x to any point within the segment [x; y], including the bounds.

$\text{[math]}$ Determine if Pig can visit the friend using teleports only, or he should use his car.

Input

The first line contains two integers n and m (1 ≤ n ≤ 100, 1 ≤ m ≤ 100) — the number of teleports and the location of the friend’s house.

The next n lines contain information about teleports.

The i-th of these lines contains two integers a_i and b_i (0 ≤ a_i ≤ b_i ≤ m), where a_i is the location of the i-th teleport, and b_i is its limit.

It is guaranteed that a_i ≥ a_i - 1 for every i (2 ≤ i ≤ n).

Output

Print “YES” if there is a path from Pig’s house to his friend’s house that uses only teleports, and “NO” otherwise.

You can print each letter in arbitrary case (upper or lower).

Examples

input

3 5

0 2

2 4

3 5

output

YES

YES

input

3 7

0 4

2 5

6 7

output

NO

Note

The first example is shown on the picture below:

$\text{[math]}$ Pig can use the first teleport from his house (point 0) to reach point 2, then using the second teleport go from point 2 to point 3, then using the third teleport go from point 3 to point 5, where his friend lives.

The second example is shown on the picture below:

$\text{[math]}$ You can see that there is no path from Pig’s house to his friend’s house that uses only teleports.

我的说明

题目大意是有一只猪去找朋友玩，可以使用传送门瞬间移动。每个传送门有一定的传送距离，可以传送至传送范围内的任意一点。判断它能否只使用传送门到朋友家。

输入数据

第一行有两个数n m，n表示传送门的个数，m表示朋友家距离原点的距离

接下来n行每行两个数 a_i b_i ，其中a_i 表示传送门的位置，b_i 表示传送门最远可以传送到的位置（两者均是相对于原点的绝对位置）， a_i 保证从小到大输入

输出数据

一行，如果可以仅通过传送门到达朋友家输出YES，如果不能输出NO

我的题解

通过贪心的方法，记录当前可到达的最远距离，如果下一个传送门的位置小于等于当前可传送到的最远距离，且该传送门可传送的距离大于当前可达到的最远距离，更新当前可达到的最远距离。

如果出现某一传送门（或朋友家）在当前可达到的最远距离之后，则说明无法仅通过传送门到达朋友家。

#include <cstdio>
#include <cstdlib>
#define max(a,b)    (((a) > (b)) ? (a) : (b))
#define min(a,b)    (((a) < (b)) ? (a) : (b))
int main() {
	int n, m;
	int r;
	r = 0;
	int q, w;
	scanf("%d %d", &n, &m);
	for (int i = 1; i <= n; i++) {
		scanf("%d %d", &q, &w);
		if (q <= r) r = max(r, w);
		else {
			printf("NO");
			return 0;
		}		
	}
	if (r>=m)
	printf("YES");
	else
	printf("NO");
	return 0;	
}

#include <cstdio>

#include <cstdlib>

#define max(a,b) (((a) > (b)) ? (a) : (b))

#define min(a,b) (((a) < (b)) ? (a) : (b))

int main() {

int n, m;

int r;

r = 0;

int q, w;

scanf("%d %d", &n, &m);

for (int i = 1; i <= n; i++) {

scanf("%d %d", &q, &w);

if (q <= r) r = max(r, w);

else {

printf("NO");

return 0;

}

if (r>=m)

printf("YES");

else

printf("NO");

return 0;

}

You are given a rooted tree with n vertices. The vertices are numbered from 1 to n, the root is the vertex number 1.

Each vertex has a color, let’s denote the color of vertex v by c_v. Initially c_v = 0.

You have to color the tree into the given colors using the smallest possible number of steps. On each step you can choose a vertex v and a color x, and then color all vectices in the subtree of v (including v itself) in color x. In other words, for every vertex u, such that the path from root to u passes through v, set c_u = x.

It is guaranteed that you have to color each vertex in a color different from 0.

You can learn what a rooted tree is using the link: https://en.wikipedia.org/wiki/Tree_(graph_theory).

Input

The first line contains a single integer n (2 ≤ n ≤ 10⁴) — the number of vertices in the tree.

The second line contains n - 1 integers p₂, p₃, …, p_n (1 ≤ p_i < i), where p_i means that there is an edge between vertices i and p_i.

The third line contains n integers c₁, c₂, …, c_n (1 ≤ c_i ≤ n), where c_i is the color you should color the i-th vertex into.

It is guaranteed that the given graph is a tree.

Output

Print a single integer — the minimum number of steps you have to perform to color the tree into given colors.

Examples

input

6
1 2 2 1 5
2 1 1 1 1 1

1 2 2 1 5

2 1 1 1 1 1

output

3

input

7
1 1 2 3 1 4
3 3 1 1 1 2 3

1 1 2 3 1 4

3 3 1 1 1 2 3

output

5

Note

The tree from the first sample is shown on the picture (numbers are vetices’ indices):

$\text{[math]}$

On first step we color all vertices in the subtree of vertex 1 into color 2 (numbers are colors):

$\text{[math]}$

On seond step we color all vertices in the subtree of vertex 5 into color 1:

$\text{[math]}$

On third step we color all vertices in the subtree of vertex 2 into color 1:

$\text{[math]}$

The tree from the second sample is shown on the picture (numbers are vetices’ indices):

$\text{[math]}$

On first step we color all vertices in the subtree of vertex 1 into color 3 (numbers are colors):

$\text{[math]}$

On second step we color all vertices in the subtree of vertex 3 into color 1:

$\text{[math]}$

On third step we color all vertices in the subtree of vertex 6 into color 2:

$\text{[math]}$

On fourth step we color all vertices in the subtree of vertex 4 into color 1:

$\text{[math]}$

On fith step we color all vertices in the subtree of vertex 7 into color 3:

$\text{[math]}$

我的理解

题目大意是给定一棵有根树，要求给每个节点染色，每次染色将某一节点及其所有子节点染为同一颜色

我的题解

显然每次染色只能影响到子节点而不能影响到父节点，所以从根节点开始向下染色，如果当前颜色与要求的颜色不一样则染为要求的颜色。进一步优化可知若节点与父节点的颜色不同则应染一次色。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#define max(a,b)    (((a) > (b)) ? (a) : (b))
#define min(a,b)    (((a) < (b)) ? (a) : (b))
int tree[10001];
int color[10001];
int main() {
	memset(tree, 0, sizeof(tree));
	memset(color, 0, sizeof(color));
	int n;
	scanf("%d", &n);
	tree[1] = 0;
	for (int i = 2; i <= n; i++) {
		scanf("%d", &tree[i]);
	}
	for (int i = 1; i <= n; i++) {
		scanf("%d", &color[i]);
	}
	int count = 0;
	for (int i = 1; i <= n; i++) {
		if (color[i] != color[tree[i]]) count++;
	}
	printf("%d", count);
	return 0;
}

#include <cstdio>

#include <cstdlib>

#include <cstring>

#define max(a,b) (((a) > (b)) ? (a) : (b))

#define min(a,b) (((a) < (b)) ? (a) : (b))

int tree[10001];

int color[10001];

int main() {

memset(tree, 0, sizeof(tree));

memset(color, 0, sizeof(color));

int n;

scanf("%d", &n);

tree[1] = 0;

for (int i = 2; i <= n; i++) {

scanf("%d", &tree[i]);

}

for (int i = 1; i <= n; i++) {

scanf("%d", &color[i]);

}

int count = 0;

for (int i = 1; i <= n; i++) {

if (color[i] != color[tree[i]]) count++;

}

printf("%d", count);

return 0;

}

以下题目我都没AC。就不增加自己的说明了

Sasha is taking part in a programming competition. In one of the problems she should check if some rooted trees are isomorphic or not. She has never seen this problem before, but, being an experienced participant, she guessed that she should match trees to some sequences and then compare these sequences instead of trees. Sasha wants to match each tree with a sequence a₀, a₁, …, a_h, where his the height of the tree, and a_i equals to the number of vertices that are at distance of i edges from root.

Unfortunately, this time Sasha’s intuition was wrong, and there could be several trees matching the same sequence. To show it, you need to write a program that, given the sequence a_i, builds two non-isomorphic rooted trees that match that sequence, or determines that there is only one such tree.

Two rooted trees are isomorphic, if you can reenumerate the vertices of the first one in such a way, that the index of the root becomes equal the index of the root of the second tree, and these two trees become equal.

The height of a rooted tree is the maximum number of edges on a path from the root to any other vertex.

Input

The first line contains a single integer h (2 ≤ h ≤ 10⁵) — the height of the tree.

The second line contains h + 1 integers — the sequence a₀, a₁, …, a_h (1 ≤ a_i ≤ 2·10⁵). The sum of all a_i does not exceed 2·10⁵. It is guaranteed that there is at least one tree matching this sequence.

Output

If there is only one tree matching this sequence, print “perfect“.

Otherwise print “ambiguous” in the first line. In the second and in the third line print descriptions of two trees in the following format: in one line print $\text{[math]}$ integers, the k-th of them should be the parent of vertex k or be equal to zero, if the k-th vertex is the root.

These treese should be non-isomorphic and should match the given sequence.

Examples

input

2
1 1 1

1 1 1

output

perfect

perfect

input

2
1 2 2

1 2 2

output

ambiguous
0 1 1 3 3
0 1 1 3 2

ambiguous

0 1 1 3 3

0 1 1 3 2

Note

The only tree in the first example and the two printed trees from the second example are shown on the picture:

$\text{[math]}$

Suppose you have two polynomials $\text{[math]}$ and $\text{[math]}$ . Then polynomial $\text{[math]}$ can be uniquely represented in the following way:

$\text{[math]}$ This can be done using long division. Here, $\text{[math]}$ denotes the degree of polynomial P(x). $\text{[math]}$ is called the remainder of division of polynomial $\text{[math]}$ by polynomial $\text{[math]}$ , it is also denoted as $\text{[math]}$ .

Since there is a way to divide polynomials with remainder, we can define Euclid’s algorithm of finding the greatest common divisor of two polynomials. The algorithm takes two polynomials $\text{[math]}$ . If the polynomial $\text{[math]}$ is zero, the result is $\text{[math]}$ , otherwise the result is the value the algorithm returns for pair $\text{[math]}$ . On each step the degree of the second argument decreases, so the algorithm works in finite number of steps. But how large that number could be? You are to answer this question.

You are given an integer n. You have to build two polynomials with degrees not greater than n, such that their coefficients are integers not exceeding 1 by their absolute value, the leading coefficients (ones with the greatest power of x) are equal to one, and the described Euclid’s algorithm performs exactly n steps finding their greatest common divisor. Moreover, the degree of the first polynomial should be greater than the degree of the second. By a step of the algorithm we mean the transition from pair $\text{[math]}$ to pair $\text{[math]}$ .

Input

You are given a single integer n (1 ≤ n ≤ 150) — the number of steps of the algorithm you need to reach.

Output

Print two polynomials in the following format.

In the first line print a single integer m (0 ≤ m ≤ n) — the degree of the polynomial.

In the second line print m + 1 integers between - 1 and 1 — the coefficients of the polynomial, from constant to leading.

The degree of the first polynomial should be greater than the degree of the second polynomial, the leading coefficients should be equal to 1. Euclid’s algorithm should perform exactly n steps when called using these polynomials.

If there is no answer for the given n, print -1.

If there are multiple answer, print any of them.

Examples

input

1

output

0 1

input

2

output

-1 0 1

0 1

Note

In the second example you can print polynomials x² - 1 and x. The sequence of transitions is

(x² - 1, x) → (x, - 1) → ( - 1, 0).There are two steps in it.

You are given an undirected graph with n vertices. There are no edge-simple cycles with the even length in it. In other words, there are no cycles of even length that pass each edge at most once. Let’s enumerate vertices from 1 to n.

You have to answer q queries. Each query is described by a segment of vertices [l; r], and you have to count the number of its subsegments [x; y] (l ≤ x ≤ y ≤ r), such that if we delete all vertices except the segment of vertices [x; y] (including x and y) and edges between them, the resulting graph is bipartite.

Input

The first line contains two integers n and m (1 ≤ n ≤ 3·10⁵, 1 ≤ m ≤ 3·10⁵) — the number of vertices and the number of edges in the graph.

The next m lines describe edges in the graph. The i-th of these lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ n; a_i ≠ b_i), denoting an edge between vertices a_i and b_i. It is guaranteed that this graph does not contain edge-simple cycles of even length.

The next line contains a single integer q (1 ≤ q ≤ 3·10⁵) — the number of queries.

The next q lines contain queries. The i-th of these lines contains two integers l_i and r_i (1 ≤ l_i ≤ r_i ≤ n) — the query parameters.

Output

Print q numbers, each in new line: the i-th of them should be the number of subsegments [x; y] (l_i ≤ x ≤ y ≤ r_i), such that the graph that only includes vertices from segment [x; y] and edges between them is bipartite.

Examples

input

6 6

1 2

2 3

3 1

4 5

5 6

6 4

1 3

4 6

1 6

output

5
5
14

input

8 9

1 2

2 3

3 1

4 5

5 6

6 7

7 8

8 4

7 2

1 8

1 4

3 8

output

27
8
19

Note

The first example is shown on the picture below:

$\text{[math]}$

For the first query, all subsegments of [1; 3], except this segment itself, are suitable.

For the first query, all subsegments of [4; 6], except this segment itself, are suitable.

For the third query, all subsegments of [1; 6] are suitable, except [1; 3], [1; 4], [1; 5], [1; 6], [2; 6], [3; 6], [4; 6].

The second example is shown on the picture below:

$\text{[math]}$

一	二	三	四	五	六	日
« 11月				1月 »
				1	2	3
4	5	6	7	8	9	10
11	12	13	14	15	16	17
18	19	20	21	22	23	24
25	26	27	28	29	30	31

A. Visiting a Friend

Input

It is guaranteed that ai ≥ ai - 1 for every i (2 ≤ i ≤ n).

Output

Examples

input

output

input

output

Note

我的说明

我的题解

B. Coloring a Tree

Input

Output

Examples

input

output

input

output

Note

我的理解

我的题解

以下题目我都没AC。就不增加自己的说明了

C. Hashing Trees

Input

Output

Examples

input

output

input

output

Note

D. GCD of Polynomials

Input

Output

Examples

input

output

input

output

Note

E. Bipartite Segments

Input

Output

Examples

input

output

input

output

Note

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It is guaranteed that a_i ≥ a_i - 1 for every i (2 ≤ i ≤ n).

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