#include<bits/stdc++.h>
#define maxn 400010
using namespace std;
char s[maxn];
int out[maxn][2], used[maxn], dl[maxn], head[maxn], nxt[maxn * 2], a[maxn * 2], edge;
void create(int u, int v)
{
	edge++; a[edge] = v; nxt[edge] = head[u]; head[u] = edge;
}
int main()
{
	int n, m;
	scanf("%d%d", &n, &m);
	scanf("%s", s + 1);
	for (int i = 1; i <= m; i++)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		create(u, v); create(v, u);
		if (s[v] == 'A') out[u][0]++; else out[u][1]++;
		if (s[u] == 'A') out[v][0]++; else out[v][1]++;
	}
	int lef = 1, righ = 0;
	for (int i = 1; i <= n; i++)
		if (out[i][0] == 0 || out[i][1] == 0) {used[i] = 1; dl[++righ] = i;}
	while (lef <= righ)
	{
		int u = dl[lef];
		for (int i = head[u]; i; i = nxt[i])
		{
			int v = a[i];
			if (s[u] == 'A') out[v][0]--; else out[v][1]--;
			if (used[v] == 0 && (out[v][0] == 0 || out[v][1] == 0))
			{
				used[v] = 1;
				dl[++righ] = v;
			}
		}
		lef++;
	}
	for (int i = 1; i <= n; i++)
		if (used[i] == 0) {puts("Yes"); return 0;}
	puts("No");
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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<bits/stdc++.h>
#define maxn 400010
using namespace std;
char s[maxn];
int out[maxn][2], used[maxn], dl[maxn], head[maxn], nxt[maxn * 2], a[maxn * 2], edge;
void create(int u, int v)
{
	edge++; a[edge] = v; nxt[edge] = head[u]; head[u] = edge;
}
int main()
{
	int n, m;
	scanf("%d%d", &n, &m);
	scanf("%s", s + 1);
	for (int i = 1; i <= m; i++)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		create(u, v); create(v, u);
		if (s[v] == 'A') out[u][0]++; else out[u][1]++;
		if (s[u] == 'A') out[v][0]++; else out[v][1]++;
	}
	int lef = 1, righ = 0;
	for (int i = 1; i <= n; i++)
		if (out[i][0] == 0 || out[i][1] == 0) {used[i] = 1; dl[++righ] = i;}
	while (lef <= righ)
	{
		int u = dl[lef];
		for (int i = head[u]; i; i = nxt[i])
		{
			int v = a[i];
			if (s[u] == 'A') out[v][0]--; else out[v][1]--;
			if (used[v] == 0 && (out[v][0] == 0 || out[v][1] == 0))
			{
				used[v] = 1;
				dl[++righ] = v;
			}
		}
		lef++;
	}
	for (int i = 1; i <= n; i++)
		if (used[i] == 0) {puts("Yes"); return 0;}
	puts("No");
	return 0;
}
```