#include <cstdio>

const int N = 2e5 + 10;
const int M = 4e5 + 10;

int hd[N], nxt[M], to[M], cnt;
int deg[N][2];
int n, m;
char col[N];
int q[N], qhd, qtl, vis[N];

inline void adde(int x, int y) {
	cnt++;
	to[cnt] = y; nxt[cnt] = hd[x]; 
	hd[x] = cnt;
}

int main() {
	scanf("%d%d", &n, &m);
	scanf("%s", col+1);
	for (int i = 1, x, y; i <= m; ++i) {
		scanf("%d%d", &x, &y);
		adde(x, y);
		adde(y, x);
		deg[x][col[y] == 'B']++;
		deg[y][col[x] == 'B']++;
	}
	for (int i = 1; i <= n; ++i) {
		if (deg[i][0] == 0 || deg[i][1] == 0) vis[q[qtl++] = i] = 1;
	}
	while (qhd != qtl) {
		int x = q[qhd++];
		// printf("%d\n", x);
		// vis[x] = 1;
		for (int i = hd[x]; i; i = nxt[i]) if (!vis[to[i]]) {
			if (--deg[to[i]][col[x] == 'B'] == 0) {
				vis[to[i]] = 1;
				q[qtl++] = to[i];
			} 
		}
	}
	if (qtl == n) puts("No");
	else puts("Yes");
}

### Prompt

In cpp, your task is to solve the following problem:

You are given an undirected graph consisting of N vertices and M edges. The vertices are numbered 1 to N, and the edges are numbered 1 to M. In addition, each vertex has a label, `A` or `B`. The label of Vertex i is s_i. Edge i bidirectionally connects vertex a_i and b_i.

The phantom thief Nusook likes to choose some vertex as the startpoint and traverse an edge zero or more times. Today, he will make a string after traveling as above, by placing the labels of the visited vertices in the order visited, beginning from the startpoint.

For example, in a graph where Vertex 1 has the label `A` and Vertex 2 has the label `B`, if Nusook travels along the path 1 \rightarrow 2 \rightarrow 1 \rightarrow 2 \rightarrow 2, the resulting string is `ABABB`.

Determine if Nusook can make all strings consisting of `A` and `B`.

Constraints

* 2 \leq N \leq 2 \times 10^{5}
* 1 \leq M \leq 2 \times 10^{5}
* |s| = N
* s_i is `A` or `B`.
* 1 \leq a_i, b_i \leq N
* The given graph may NOT be simple or connected.

Input

Input is given from Standard Input in the following format:


N M
s
a_1 b_1
:
a_{M} b_{M}


Output

If Nusook can make all strings consisting of `A` and `B`, print `Yes`; otherwise, print `No`.

Examples

Input

2 3
AB
1 1
1 2
2 2


Output

Yes


Input

4 3
ABAB
1 2
2 3
3 1


Output

No


Input

13 23
ABAAAABBBBAAB
7 1
10 6
1 11
2 10
2 8
2 11
11 12
8 3
7 12
11 2
13 13
11 9
4 1
9 7
9 6
8 13
8 6
4 10
8 7
4 3
2 1
8 12
6 9


Output

Yes


Input

13 17
BBABBBAABABBA
7 1
7 9
11 12
3 9
11 9
2 1
11 5
12 11
10 8
1 11
1 8
7 7
9 10
8 8
8 12
6 2
13 11


Output

No

### Response

```cpp
#include <cstdio>

const int N = 2e5 + 10;
const int M = 4e5 + 10;

int hd[N], nxt[M], to[M], cnt;
int deg[N][2];
int n, m;
char col[N];
int q[N], qhd, qtl, vis[N];

inline void adde(int x, int y) {
	cnt++;
	to[cnt] = y; nxt[cnt] = hd[x]; 
	hd[x] = cnt;
}

int main() {
	scanf("%d%d", &n, &m);
	scanf("%s", col+1);
	for (int i = 1, x, y; i <= m; ++i) {
		scanf("%d%d", &x, &y);
		adde(x, y);
		adde(y, x);
		deg[x][col[y] == 'B']++;
		deg[y][col[x] == 'B']++;
	}
	for (int i = 1; i <= n; ++i) {
		if (deg[i][0] == 0 || deg[i][1] == 0) vis[q[qtl++] = i] = 1;
	}
	while (qhd != qtl) {
		int x = q[qhd++];
		// printf("%d\n", x);
		// vis[x] = 1;
		for (int i = hd[x]; i; i = nxt[i]) if (!vis[to[i]]) {
			if (--deg[to[i]][col[x] == 'B'] == 0) {
				vis[to[i]] = 1;
				q[qtl++] = to[i];
			} 
		}
	}
	if (qtl == n) puts("No");
	else puts("Yes");
}

```