#include <bits/stdc++.h>
using namespace std;

int fst[400010], nxt[400010], lst[400010], des[400010];
int q[400010], s[400010][2], vis[400010], cnt = 0;
char st[400010];

void add(int u, int v) {
	if (!fst[u]) fst[u] = ++cnt;
	else nxt[lst[u]] = ++cnt;
	lst[u] = cnt, des[cnt] = v;
}

int main() {

	int n, m; scanf("%d%d", &n, &m);
	scanf("%s", st + 1);
	for (int i = 1; i <= m; ++i) {
		int u, v; scanf("%d%d", &u, &v);
		add(u, v), add(v, u);
		s[u][st[v] - 'A'] ++;
		s[v][st[u] - 'A'] ++;
	}
	int sum = 0, r = 0, l = 0;
	for (int i = 1; i <= n; ++i)
		if (!s[i][0] || !s[i][1]) 
			q[++r] = i, vis[i] = 1, sum ++;
	while (l <= r) {
		for (int i = fst[q[l]]; i; i = nxt[i])
			if (!vis[des[i]]) {
				s[des[i]][st[q[l]] - 'A'] --;
				if (!s[des[i]][0] || !s[des[i]][1]) {
					vis[des[i]] = 1;
					q[++r] = des[i];
					sum ++;
				}
			}
		++ l;
	}
	if (sum < n) puts("Yes");
	else puts("No");
	return 0;

}
### Prompt

Generate 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 <bits/stdc++.h>
using namespace std;

int fst[400010], nxt[400010], lst[400010], des[400010];
int q[400010], s[400010][2], vis[400010], cnt = 0;
char st[400010];

void add(int u, int v) {
	if (!fst[u]) fst[u] = ++cnt;
	else nxt[lst[u]] = ++cnt;
	lst[u] = cnt, des[cnt] = v;
}

int main() {

	int n, m; scanf("%d%d", &n, &m);
	scanf("%s", st + 1);
	for (int i = 1; i <= m; ++i) {
		int u, v; scanf("%d%d", &u, &v);
		add(u, v), add(v, u);
		s[u][st[v] - 'A'] ++;
		s[v][st[u] - 'A'] ++;
	}
	int sum = 0, r = 0, l = 0;
	for (int i = 1; i <= n; ++i)
		if (!s[i][0] || !s[i][1]) 
			q[++r] = i, vis[i] = 1, sum ++;
	while (l <= r) {
		for (int i = fst[q[l]]; i; i = nxt[i])
			if (!vis[des[i]]) {
				s[des[i]][st[q[l]] - 'A'] --;
				if (!s[des[i]][0] || !s[des[i]][1]) {
					vis[des[i]] = 1;
					q[++r] = des[i];
					sum ++;
				}
			}
		++ l;
	}
	if (sum < n) puts("Yes");
	else puts("No");
	return 0;

}
```