#include <cstdio>
#define V_MAX 200000
#define E_MAX 200000
struct edg
{
	int v; edg * e;
} ES[E_MAX * 2], * ER = ES, * EA[V_MAX + 1];
inline void edg_ins(int u, int v)
{
	*ER = (edg) {v, EA[u]}, EA[u] = ER++;
	*ER = (edg) {u, EA[v]}, EA[v] = ER++;
}
int V, E, u, v, k, d[V_MAX + 1][2], Ql, Qr, Q[V_MAX + 1];
char s[V_MAX + 2];
int main()
{
	scanf("%d %d", &V, &E);
	scanf(" %s", s + 1);
	while (E--)
		scanf("%d %d", &u, &v), edg_ins(u, v), ++d[u][s[v] - 'A'], ++d[v][s[u] - 'A'];
	for (u = 1; u <= V; ++u)
		if (!d[u][0] || !d[u][1])
			Q[++Qr] = u;
	while (Ql < Qr)
	{
		u = Q[++Ql], k = s[u] - 'A', --V;
		for (edg * e = EA[u]; e; e = e->e)
			if (!--d[e->v][k] && d[e->v][k ^ 1])
				Q[++Qr] = e->v;
	}
	puts(V ? "Yes" : "No");
	return 0;
}
### Prompt

Your challenge is to write a Cpp solution to 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>
#define V_MAX 200000
#define E_MAX 200000
struct edg
{
	int v; edg * e;
} ES[E_MAX * 2], * ER = ES, * EA[V_MAX + 1];
inline void edg_ins(int u, int v)
{
	*ER = (edg) {v, EA[u]}, EA[u] = ER++;
	*ER = (edg) {u, EA[v]}, EA[v] = ER++;
}
int V, E, u, v, k, d[V_MAX + 1][2], Ql, Qr, Q[V_MAX + 1];
char s[V_MAX + 2];
int main()
{
	scanf("%d %d", &V, &E);
	scanf(" %s", s + 1);
	while (E--)
		scanf("%d %d", &u, &v), edg_ins(u, v), ++d[u][s[v] - 'A'], ++d[v][s[u] - 'A'];
	for (u = 1; u <= V; ++u)
		if (!d[u][0] || !d[u][1])
			Q[++Qr] = u;
	while (Ql < Qr)
	{
		u = Q[++Ql], k = s[u] - 'A', --V;
		for (edg * e = EA[u]; e; e = e->e)
			if (!--d[e->v][k] && d[e->v][k ^ 1])
				Q[++Qr] = e->v;
	}
	puts(V ? "Yes" : "No");
	return 0;
}
```