#include <bits/stdc++.h>
#define MAX_N 200000
using namespace std;
template <class T> inline void read(T &x) {
	x = 0; int c = getchar(), f = 1;
	for (; !isdigit(c); c = getchar()) if (c == 45) f = -1;
	for (; isdigit(c); c = getchar()) (x *= 10) += f*(c-'0');
}
char s[MAX_N+5]; bool mrk[MAX_N+5];
int n, m, c[MAX_N+5], d[MAX_N+5][2];
vector <int> G[MAX_N+5]; queue <int> que;
int main() {
	read(n), read(m), scanf("%s", s+1);
	for (int i = 1; i <= n; i++) c[i] = s[i] == 'B';
	for (int i = 1, u, v; i <= m; i++)
		read(u), read(v), 
		G[u].push_back(v), d[v][c[u]]++, 
		G[v].push_back(u), d[u][c[v]]++;
	for (int i = 1; i <= n; i++)
		if (!d[i][0] || !d[i][1])
			que.push(i), mrk[i] = true;
	while (!que.empty()) {
		int u = que.front(); que.pop();
		for (int v : G[u])
			if (!mrk[v] && !--d[v][c[u]])
				que.push(v), mrk[v] = true;
	}
	for (int i = 1; i <= n; i++)
		if (!mrk[i]) return puts("Yes"), 0;
	return puts("No"), 0;
}
### Prompt

Create a solution in CPP for 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>
#define MAX_N 200000
using namespace std;
template <class T> inline void read(T &x) {
	x = 0; int c = getchar(), f = 1;
	for (; !isdigit(c); c = getchar()) if (c == 45) f = -1;
	for (; isdigit(c); c = getchar()) (x *= 10) += f*(c-'0');
}
char s[MAX_N+5]; bool mrk[MAX_N+5];
int n, m, c[MAX_N+5], d[MAX_N+5][2];
vector <int> G[MAX_N+5]; queue <int> que;
int main() {
	read(n), read(m), scanf("%s", s+1);
	for (int i = 1; i <= n; i++) c[i] = s[i] == 'B';
	for (int i = 1, u, v; i <= m; i++)
		read(u), read(v), 
		G[u].push_back(v), d[v][c[u]]++, 
		G[v].push_back(u), d[u][c[v]]++;
	for (int i = 1; i <= n; i++)
		if (!d[i][0] || !d[i][1])
			que.push(i), mrk[i] = true;
	while (!que.empty()) {
		int u = que.front(); que.pop();
		for (int v : G[u])
			if (!mrk[v] && !--d[v][c[u]])
				que.push(v), mrk[v] = true;
	}
	for (int i = 1; i <= n; i++)
		if (!mrk[i]) return puts("Yes"), 0;
	return puts("No"), 0;
}
```