#include <cstdio>
#include <cstring>
 
int c[201000]; char s[201000];
int st[201000], top = 0, inq[201000][2], T[201000];
int h[201000], nxt[401000], to[401000], K = 0;
bool used[201000][2];
int flag = 0;

void ins(int u, int v) {nxt[++K] = h[u]; h[u] = K; to[K] = v;}

void dfs(int x, int t)
{
	if(inq[x][t]) {flag = 1; return;}
	if(used[x][t]) return;
	st[++top] = x; T[top] = t; 
	inq[st[top]][T[top]] = 1;
	used[x][t] = 1;
	for(int i = h[x]; i; i = nxt[i])
	{
		if(flag) return;
		if(c[to[i]] == (t ^ 1))
		{
			dfs(to[i], c[x]);
		}
	}
	inq[st[top]][T[top]] = 0;
	top--;
}
 
int main()
{
	int n, m; scanf("%d%d", &n, &m);
	scanf("%s", s + 1);
	for(int i = 1; i <= n; i++) c[i] = (s[i] == 'A' ? 1 : 0);
	for(int i = 1; i <= m; i++)
	{
		int u, v; scanf("%d%d", &u, &v);
		ins(u, v); ins(v, u);
	}
	for(int i = 1; i <= n; i++)
	{
		if(!used[i][0])
		{
			dfs(i, 0);
			if(flag) return 0 * puts("Yes");
		}
		if(!used[i][1])
		{
			dfs(i, 1);
			if(flag) return 0 * puts("Yes");
		}
	}
	puts("No");
	return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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>
#include <cstring>
 
int c[201000]; char s[201000];
int st[201000], top = 0, inq[201000][2], T[201000];
int h[201000], nxt[401000], to[401000], K = 0;
bool used[201000][2];
int flag = 0;

void ins(int u, int v) {nxt[++K] = h[u]; h[u] = K; to[K] = v;}

void dfs(int x, int t)
{
	if(inq[x][t]) {flag = 1; return;}
	if(used[x][t]) return;
	st[++top] = x; T[top] = t; 
	inq[st[top]][T[top]] = 1;
	used[x][t] = 1;
	for(int i = h[x]; i; i = nxt[i])
	{
		if(flag) return;
		if(c[to[i]] == (t ^ 1))
		{
			dfs(to[i], c[x]);
		}
	}
	inq[st[top]][T[top]] = 0;
	top--;
}
 
int main()
{
	int n, m; scanf("%d%d", &n, &m);
	scanf("%s", s + 1);
	for(int i = 1; i <= n; i++) c[i] = (s[i] == 'A' ? 1 : 0);
	for(int i = 1; i <= m; i++)
	{
		int u, v; scanf("%d%d", &u, &v);
		ins(u, v); ins(v, u);
	}
	for(int i = 1; i <= n; i++)
	{
		if(!used[i][0])
		{
			dfs(i, 0);
			if(flag) return 0 * puts("Yes");
		}
		if(!used[i][1])
		{
			dfs(i, 1);
			if(flag) return 0 * puts("Yes");
		}
	}
	puts("No");
	return 0;
}
```