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

typedef long long ll;
const int maxn = 200005;

inline int gi()
{
	char c = getchar();
	while (c < '0' || c > '9') c = getchar();
	int sum = 0;
	while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
	return sum;
}

int n, m;
char S[maxn];

struct edge
{
	int to, next;
} e[maxn * 2];
int h[maxn], deg[maxn][2], vis[maxn], tot;
int s, t, q[maxn];

void add(int u, int v)
{
	e[++tot] = (edge) {v, h[u]}; h[u] = tot;
	e[++tot] = (edge) {u, h[v]}; h[v] = tot;
	++deg[u][S[v] - 'A']; ++deg[v][S[u] - 'A'];
}

int topo()
{
	for (int i = 1; i <= n; ++i)
		if (!deg[i][0] || !deg[i][1]) q[++t] = i, vis[i] = 1;
	int cnt = 0;
	while (s < t) {
		int u = q[++s];
		++cnt;
		for (int i = h[u], v; v = e[i].to, i; i = e[i].next)
			if (!vis[v] && !(--deg[v][S[u] - 'A'])) vis[q[++t] = v] = 1;
	}
	return cnt;
}

int main()
{
	scanf("%d%d%s", &n, &m, S + 1);
	for (int i = 1; i <= m; ++i) add(gi(), gi());
	
	puts(topo() == n ? "No" : "Yes");
	
	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;

typedef long long ll;
const int maxn = 200005;

inline int gi()
{
	char c = getchar();
	while (c < '0' || c > '9') c = getchar();
	int sum = 0;
	while ('0' <= c && c <= '9') sum = sum * 10 + c - 48, c = getchar();
	return sum;
}

int n, m;
char S[maxn];

struct edge
{
	int to, next;
} e[maxn * 2];
int h[maxn], deg[maxn][2], vis[maxn], tot;
int s, t, q[maxn];

void add(int u, int v)
{
	e[++tot] = (edge) {v, h[u]}; h[u] = tot;
	e[++tot] = (edge) {u, h[v]}; h[v] = tot;
	++deg[u][S[v] - 'A']; ++deg[v][S[u] - 'A'];
}

int topo()
{
	for (int i = 1; i <= n; ++i)
		if (!deg[i][0] || !deg[i][1]) q[++t] = i, vis[i] = 1;
	int cnt = 0;
	while (s < t) {
		int u = q[++s];
		++cnt;
		for (int i = h[u], v; v = e[i].to, i; i = e[i].next)
			if (!vis[v] && !(--deg[v][S[u] - 'A'])) vis[q[++t] = v] = 1;
	}
	return cnt;
}

int main()
{
	scanf("%d%d%s", &n, &m, S + 1);
	for (int i = 1; i <= m; ++i) add(gi(), gi());
	
	puts(topo() == n ? "No" : "Yes");
	
	return 0;
}

```