#include<cstdio>
#include<algorithm>
#include<vector>
#define N_ 201000
using namespace std;
int n, m, w[N_], v[N_], head, tail, Deg[N_][2];
char p[201000];
vector<int>E[N_][2];
int Q[N_*2];
void Add(int a, int b) {
	if (p[b] == 'A')E[a][0].push_back(b);
	else E[a][1].push_back(b);
}
int main() {
	int i, a, b, j;
	scanf("%d%d", &n, &m);
	scanf("%s", p + 1);
	for (i = 0; i < m; i++) {
		scanf("%d%d", &a, &b);
		Add(a, b);
		Add(b, a);
	}
	for (i = 1; i <= n; i++) {
		for (j = 0; j < 2; j++) {
			Deg[i][j] = E[i][j].size();
			if (E[i][j].size() == 0 && !v[i]) {
				Q[++tail] = i;
				v[i] = 1;
			}
		}
	}
	while (head < tail) {
		int x = Q[++head];
		int ck = p[x] - 'A';
		for (auto &y : E[x][0]) {
			Deg[y][ck]--;
			if (!Deg[y][ck] && !v[y]) {
				Q[++tail] = y;
				v[y] = 1;
			}
		}
		for (auto &y : E[x][1]) {
			Deg[y][ck]--;
			if (!Deg[y][ck] && !v[y]) {
				Q[++tail] = y;
				v[y] = 1;
			}
		}
	}
	if (tail == 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>
#include<algorithm>
#include<vector>
#define N_ 201000
using namespace std;
int n, m, w[N_], v[N_], head, tail, Deg[N_][2];
char p[201000];
vector<int>E[N_][2];
int Q[N_*2];
void Add(int a, int b) {
	if (p[b] == 'A')E[a][0].push_back(b);
	else E[a][1].push_back(b);
}
int main() {
	int i, a, b, j;
	scanf("%d%d", &n, &m);
	scanf("%s", p + 1);
	for (i = 0; i < m; i++) {
		scanf("%d%d", &a, &b);
		Add(a, b);
		Add(b, a);
	}
	for (i = 1; i <= n; i++) {
		for (j = 0; j < 2; j++) {
			Deg[i][j] = E[i][j].size();
			if (E[i][j].size() == 0 && !v[i]) {
				Q[++tail] = i;
				v[i] = 1;
			}
		}
	}
	while (head < tail) {
		int x = Q[++head];
		int ck = p[x] - 'A';
		for (auto &y : E[x][0]) {
			Deg[y][ck]--;
			if (!Deg[y][ck] && !v[y]) {
				Q[++tail] = y;
				v[y] = 1;
			}
		}
		for (auto &y : E[x][1]) {
			Deg[y][ck]--;
			if (!Deg[y][ck] && !v[y]) {
				Q[++tail] = y;
				v[y] = 1;
			}
		}
	}
	if (tail == n)puts("No");
	else puts("Yes");
}
```