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 ll = long long;
using ld = long double;
using V = vector<int>;
using Vll = vector<ll>;
using Vld = vector<ld>;
using Vbo = vector<bool>;
using VV = vector<V>;
using VVll = vector<Vll>;
using VVld = vector<Vld>;
using VVbo = vector<Vbo>;
using P = pair<int, int>;
using Pll = pair<ll, ll>;
using Pld = pair<ld, ld>;
#define rep2(i, m, n) for(ll i=int(m); i<int(n); ++i)
#define drep2(i, m, n) for(ll i=int(m)-1; i>=int(n); --i)
#define rep(i, n) rep2(i, 0, n)
#define drep(i, n) drep2(i, n, 0)
#define all(a) a.begin(), a.end()
struct fast_ios { fast_ios(){ cin.tie(0); ios::sync_with_stdio(false); cout << fixed << setprecision(20); }; } fast_ios_;
template<typename T> inline bool chmin(T &a, T b) { if (a > b) { a = b; return true; } return false; }
template<typename T> inline bool chmax(T &a, T b) { if (a < b) { a = b; return true; } return false; }
template<typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p) { is >> p.first >> p.second; return is; }
template<typename T1, typename T2> ostream &operator<<(ostream &os, const pair<T1, T2> &p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
template<typename T> istream &operator>>(istream &is, vector<T> &v) { for (auto &e : v) is >> e; return is; }
template<typename T> ostream &operator<<(ostream &os, const vector<T> &v) { for (auto &e : v) os << e << " "; return os; }
template<typename T> inline int count_between(vector<T> &a, T l, T r) { return lower_bound(all(a), r) - lower_bound(all(a), l); } // [l, r)
inline int Log2(ll x) { int k; for (k = 0; x > 0; ++k) x >>= 1; return k; } // number of binary digits
const int INF  = 1<<30;
const ll INFll = 1ll<<62;
const ld EPS   = 1e-10;
const int MOD  = int(1e9)+7;


int main() {
  ll n, m; cin >> n >> m;
  string s; cin >> s;

  VV G(n);
  rep(i, m) {
    int u, v; cin >> u >> v;
    --u; --v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  rep(i, n) {
    sort(all(G[i]));
    G[i].erase(unique(all(G[i])), G[i].end());
  }

  VV deg(n, V(2));
  
  rep(u, n) for (auto v : G[u]) {
    int k = s[v] - 'A';
    ++deg[u][k];
  }

  int rmv = 0;
  queue<int> que;
  rep(i, n) if (deg[i][0] == 0 || deg[i][1] == 0) que.push(i);

  while (!que.empty()) {
    int u = que.front(); que.pop();
    int k = s[u] - 'A';
    ++rmv;
    for (auto v : G[u]) {
      --deg[v][k];
      if (deg[v][k] == 0) que.push(v);
    }
  }

  if (rmv < n) cout << "Yes" << "\n";
  else cout << "No" << "\n";
  return 0;
}