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>
#include <cstdlib>
#include <queue>
#define maxn 200005

using namespace std;

struct node{
	int to, next;
}e[maxn << 1];
int head[maxn], tot = 0;
void addedge(int u, int v){e[++tot].to = v, e[tot].next = head[u]; head[u] = tot;}

int n, m, Index[maxn][2];
char str[maxn];

bool vis[maxn];

void topo(){
	queue <int> q; while (!q.empty()) q.pop();
	
	for (int i = 1; i <= n; i++)
		if (!vis[i] && (!Index[i][0] || !Index[i][1])) q.push(i);
	
	while (!q.empty()){
		int cur = q.front(); q.pop();
		vis[cur] = true;
		
		for (int i = head[cur]; i; i = e[i].next){
			if (!--Index[e[i].to][str[cur] - 'A']){
				q.push(e[i].to);
			}
		}
	}	
}

int main(){
	scanf("%d%d%s", &n, &m, str + 1);
	
	while (m--){
		int u, v; scanf("%d%d", &u, &v);
		addedge(u, v);
		addedge(v, u);
		Index[v][str[u] - 'A']++;
		Index[u][str[v] - 'A']++;
	}
	
	topo();
	
	for (int i = 1; i <= n; i++)
		if (!vis[i]){
			puts("Yes");
			return 0;
		}
	
	puts("No");
	return 0;
}