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 <bits/stdc++.h>
using namespace std;

using i64=int64_t;

#define rep(i,x,y) for(i64 i=i64(x),i##_max_for_repmacro=i64(y); i<i##_max_for_repmacro; ++i)
#define debug(x) #x << "=" << (x)
 
#ifdef DEBUG
#define _GLIBCXX_DEBUG
#define print(x) std::cerr << debug(x) << " (L:" << __LINE__ << ")" << std::endl
#else
#define print(x)
#endif
 
const int inf=1.01e9;
const i64 inf64=4.01e18;
const double eps=1e-9;
 
template <typename T> ostream &operator<<(ostream &os, const vector<T> &vec){
    os << "[";
    for (const auto &v : vec) {
        os << v << ",";
    }
    os << "]";
    return os;
}

void solve(){
    int N,M;
    string s;
    cin >> N >> M >> s;
    vector<set<int>> graph(N);
    rep(i,0,M){
        int a,b;
        cin >> a >> b;
        --a;
        --b;
        graph[a].insert(b);
        graph[b].insert(a);
    }

    vector<bool> erased(N);
    set<int> st;
    rep(i,0,N){
        bool a=false,b=false;
        for(int j:graph[i]){
            if(s[j]=='A') a=true;
            else b=true;
        }
        if(!a or !b){
            erased[i]=true;
            st.insert(i);
        }
    }

    while(!st.empty()){
        int u=*st.begin();
        st.erase(u);
        for(int v:graph[u])
        {
            graph[v].erase(u);
        }
        for(int v:graph[u]){
            if(erased[v]) continue;
            bool a=false,b=false;
            for(int w:graph[v]){
                if(s[w]=='A') a=true;
                else b=true;
                if(a and b) break;
            }
            if(!a or !b){
                erased[v]=true;
                st.insert(v);
            }
        }
    }

    int cnt=0;
    rep(i,0,N) if(erased[i]) ++cnt;
    cout << (cnt==N?"No":"Yes") << endl;
}

int main(){
    std::cin.tie(0);
    std::ios::sync_with_stdio(false);
    cout.setf(ios::fixed);
    cout.precision(16);
    solve();
    return 0;
}