#include <cstdio>

#define maxn 200010
struct Edge {
	Edge *next;
	int to;
} *last[maxn], e[maxn << 1], *ecnt = e;
char s[maxn];
int cnt[maxn][2];
inline void link(int a, int b)
{
	++cnt[b][s[a] == 'B'];
	*++ecnt = (Edge) {last[a], b}; last[a] = ecnt;
}
bool del[maxn];
int q[maxn];
int main()
{
	int n, m; scanf("%d%d", &n, &m); scanf("%s", s + 1);
	for (int i = 1; i <= m; ++i)
	{
		int a, b; scanf("%d%d", &a, &b);
		link(a, b); link(b, a);
	}
	int head = 0, tail = 0;
	for (int i = 1; i <= n; ++i) if (!cnt[i][0] || !cnt[i][1]) del[q[++tail] = i] = 1;
	while (head < tail)
	{
		int now = q[++head];
		for (Edge *iter = last[now]; iter; iter = iter -> next)
		{
			--cnt[iter -> to][s[now] == 'B'];
			if (!del[iter -> to] && !cnt[iter -> to][s[now] == 'B']) del[q[++tail] = iter -> to] = 1;
		}
	}
	puts(tail == n ? "No" : "Yes");
	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 <cstdio>

#define maxn 200010
struct Edge {
	Edge *next;
	int to;
} *last[maxn], e[maxn << 1], *ecnt = e;
char s[maxn];
int cnt[maxn][2];
inline void link(int a, int b)
{
	++cnt[b][s[a] == 'B'];
	*++ecnt = (Edge) {last[a], b}; last[a] = ecnt;
}
bool del[maxn];
int q[maxn];
int main()
{
	int n, m; scanf("%d%d", &n, &m); scanf("%s", s + 1);
	for (int i = 1; i <= m; ++i)
	{
		int a, b; scanf("%d%d", &a, &b);
		link(a, b); link(b, a);
	}
	int head = 0, tail = 0;
	for (int i = 1; i <= n; ++i) if (!cnt[i][0] || !cnt[i][1]) del[q[++tail] = i] = 1;
	while (head < tail)
	{
		int now = q[++head];
		for (Edge *iter = last[now]; iter; iter = iter -> next)
		{
			--cnt[iter -> to][s[now] == 'B'];
			if (!del[iter -> to] && !cnt[iter -> to][s[now] == 'B']) del[q[++tail] = iter -> to] = 1;
		}
	}
	puts(tail == n ? "No" : "Yes");
	return 0;
}
```