#include <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;

int n, m, f[200200], u[200200], v[200200], a[200200], b[200200];
char s[200200];
bool del[200200];

queue<int> Q;
struct edge {int to;  edge *nxt;} *head[200200], e[400400];

int main() {
	scanf("%d%d%s",&n,&m,s+1);
	for (int i=1,u,v;i<=m;i++) {
		scanf("%d%d",&u,&v);
		if (u != v) {
			head[u] = &(e[i*2-2] = (edge) {v,head[u]});
			head[v] = &(e[i*2-1] = (edge) {u,head[v]});
		}
		if (s[v] == 'A') ++a[u];  else ++b[u];
		if (s[u] == 'A') ++a[v];  else ++b[v];
	}
	for (int i=1;i<=n;i++) if (a[i] == 0 || b[i] == 0) {del[i] = 1;  Q.push(i);}
	while (!Q.empty()) {
		int u = Q.front();  Q.pop();
		for (edge *p=head[u];p;p=p->nxt) if (!del[p->to]) {
			if (s[u] == 'A') --a[p->to];  else --b[p->to];
			if (a[p->to] == 0 || b[p->to] == 0) {del[p->to] = 1;  Q.push(p->to);}
		}
	}
	for (int i=1;i<=n;i++) if (!del[i]) {
		for (edge *p=head[i];p;p=p->nxt) if (!del[p->to] && s[i] != s[p->to]) {
			puts("Yes");
			return 0;
		} 
	}
	puts("No");
}
### 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 <bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;

int n, m, f[200200], u[200200], v[200200], a[200200], b[200200];
char s[200200];
bool del[200200];

queue<int> Q;
struct edge {int to;  edge *nxt;} *head[200200], e[400400];

int main() {
	scanf("%d%d%s",&n,&m,s+1);
	for (int i=1,u,v;i<=m;i++) {
		scanf("%d%d",&u,&v);
		if (u != v) {
			head[u] = &(e[i*2-2] = (edge) {v,head[u]});
			head[v] = &(e[i*2-1] = (edge) {u,head[v]});
		}
		if (s[v] == 'A') ++a[u];  else ++b[u];
		if (s[u] == 'A') ++a[v];  else ++b[v];
	}
	for (int i=1;i<=n;i++) if (a[i] == 0 || b[i] == 0) {del[i] = 1;  Q.push(i);}
	while (!Q.empty()) {
		int u = Q.front();  Q.pop();
		for (edge *p=head[u];p;p=p->nxt) if (!del[p->to]) {
			if (s[u] == 'A') --a[p->to];  else --b[p->to];
			if (a[p->to] == 0 || b[p->to] == 0) {del[p->to] = 1;  Q.push(p->to);}
		}
	}
	for (int i=1;i<=n;i++) if (!del[i]) {
		for (edge *p=head[i];p;p=p->nxt) if (!del[p->to] && s[i] != s[p->to]) {
			puts("Yes");
			return 0;
		} 
	}
	puts("No");
}
```