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: Accepted)

#include <stdio.h>
#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <cmath>
#include <string>
#include <cstring>

#define sz(v) ((int)((v).size()))
#define all(v) (v).begin(), (v).end()

using namespace std;

typedef long long ll;
typedef pair<int,int> pii;

const int INF = 0x3c3c3c3c;
const ll LINF = 1ll*INF*INF*2;

const int N = 200011;
int n,m;
char input[N];
bool ab[N];
vector<int> v[N][2];

bool imp[N];

int cnt;
int visited[N];
bool dfs(bool prev, int x) {
	if(imp[x]) return false;
	if(visited[x] == cnt) return true;
	visited[x] = cnt;

	bool cur = ab[x];
	
	for(int i=0; i<sz(v[x][!prev]); i++) {
		if(dfs(cur, v[x][!prev][i])) return true;
	}

	imp[x] = true;
	return false;
}

int main(void) {
	scanf("%d%d", &n, &m);
	scanf("%s", input);
	for(int i=0; i<n; i++) ab[i+1] = (input[i]=='A');

	int x, y;
	for(int i=0; i<m; i++) {
		scanf("%d%d", &x, &y);
		v[x][ab[y]].push_back(y);
		v[y][ab[x]].push_back(x);
	}

	for(int i=1; i<=n; i++) {
		cnt++;

		if(dfs(0,i) && dfs(1,i)) {
			visited[i] = 0;
			if(dfs(1,i)) {
				printf("Yes");
				return 0;
			}
		}
	}

	printf("No");
}