Problem:
Score : 900 points

Problem Statement

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.

Sample Input 1

2 3
AB
1 1
1 2
2 2

Sample Output 1

Yes

Since Nusook can visit Vertex 1 and Vertex 2 in any way he likes, he can make all strings consisting of A and B.

Sample Input 2

4 3
ABAB
1 2
2 3
3 1

Sample Output 2

No

For example, Nusook cannot make BB.

The given graph may not be connected.

Sample Input 3

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

Sample Output 3

Yes

Sample Input 4

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

Sample Output 4

No

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <cstdio>

const int N = 2e5 + 10;
const int M = 4e5 + 10;
const char S[] = "AABBA";

int hd[N], nxt[M], to[M], cnt;
char col[N];
int n, m, flag, pos;

inline void adde(int x, int y) {
	cnt++;
	nxt[cnt] = hd[x];
	to[cnt] = y;
	hd[x] = cnt;
}

inline void dfs(int x, int c) {
	// if (pos == 1) printf("%d %d\n", x, c);
	if (flag) return;
	if (c == 5) {
		if (x == pos) flag = 1;
		return;
	}
	for (int i = hd[x]; i; i = nxt[i]) {
		// printf("%c %c\n", col[to[i]], S[c]);
		if (col[to[i]] == S[c])
			dfs(to[i], c+1);
	}
}

int main() {
	scanf("%d%d", &n, &m);
	scanf("%s", col+1);
	for (int i = 1, x, y; i <= m; ++i) {
		scanf("%d%d", &x, &y);
		adde(x, y);
		adde(y, x);
	}
	for (int i = 1; i <= n; ++i) {
		if (col[i] == 'A') {
			pos = i;
			dfs(i, 1);
		} 
		if (flag) break;
	}
	printf(flag ? "Yes\n" : "No\n");
	return 0;
}