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 <bits/stdc++.h>

using namespace std;

using int64 = long long;
const int64 INF = 1LL << 60;

const string T = "AABB";
vector< int > g[200000];
int v[200000][4];
bool visited[200000];
string S;

void dfs(int idx, int type) {
  if(type == 0) visited[idx] = true;
  for(auto &to : g[idx]) {
    int nxt = (type + 1) % 4;
    if(S[to] == T[nxt] && nxt == 0 && v[to][nxt]) {
      cout << "Yes\n";
      exit(0);
    }
    if(S[to] == T[nxt] && !v[to][nxt]) {
      v[to][nxt] = true;
      dfs(to, nxt);
      v[to][nxt] = false;
    }
  }
}

int main() {
  int N, M;
  cin >> N >> M;
  cin >> S;
  for(int i = 0; i < M; i++) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  queue< pair< int, int > > que;
  for(int i = 0; i < S.size(); i++) {
    if(visited[i]) continue;
    if(S[i] == 'A') {
      v[i][0] = 1;
      dfs(i, 0);
      v[i][0] = 0;
      break;
    }
  }
  cout << "No\n";
}