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 lint = long long int;
template<class T = int> using V = vector<T>;
template<class T = int> using VV = V< V<T> >;
template<class T> void assign(V<T>& v, int n, const T& a = T()) { v.assign(n, a); }
template<class T, class... U> void assign(V<T>& v, int n, const U&... u) { v.resize(n); for (auto&& i : v) assign(i, u...); }

bool rec(int v, V<>& vis, const string& s, const VV<>& g) {
  vis[v] = 1;
  bool fa = false, fb = false;
  for (int i : g[v]) if (vis[i] == 1) {
    if (s[i] == 'A') fa = true;
    else fb = true;
  }
  if (!fa) for (int i : g[v]) if (s[i] == 'A') {
    if (rec(i, vis, s, g)) {
      fa = true;
      break;
    }
  }
  if (!fb) for (int i : g[v]) if (s[i] == 'B') {
    if (rec(i, vis, s, g)) {
      fb = true;
      break;
    }
  }
  return vis[v] = fa and fb;
}

int main() {
  cin.tie(NULL); ios::sync_with_stdio(false);
  int n, m; cin >> n >> m;
  string s; cin >> s;
  VV<> g(n);
  for (int i = 0; i < m; i++) {
    int a, b; cin >> a >> b, a--, b--;
    g[a].push_back(b);
    if (a != b) g[b].push_back(a);
  }
  V<> vis(n, -1);
  bool ga = false, gb = false;
  for (int i = 0; i < n; i++) if (vis[i] and s[i] == 'A') {
    if (rec(i, vis, s, g)) ga = true;
  }
  vis.assign(n, -1);
  for (int i = 0; i < n; i++) if (vis[i] and s[i] == 'B') {
    if (rec(i, vis, s,g )) gb = true;
  }
  cout << (ga and gb ? "Yes" : "No") << '\n';
}