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 <cstdio>
#include <iostream>
#include <string>
#include <vector>
#include <sstream>
#include <map>
#include <set>
#include <queue>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <typeinfo>
#include <numeric>
#include <functional>
#include <unordered_map>
#include <bitset>
#include <stack>


using namespace std;
using ll = long long;
using ull = unsigned long long;

const ll INF = 1e18;
const ll MOD = 1e9 + 7;

#define REP(i, n) for(int i = 0; i < n; i++)



vector<vector<int>> g;
int goal;
string s;
vector<bool> used, used2;

bool dfs1(int &p);

bool dfs4(int &p){
    bool res = false;
    for(auto &x : g[p]){
        if(s[x] == 'A'){
            if(goal == x) res = true;
            else if(!used[x]){
                used[p] = true;
                res |= dfs1(x);
            }
        }
        if(res){
            // cout << x << ' ';
            break;
        }
    }
    return res;
}

bool dfs3(int &p){
    bool res = false;
    for(auto &x : g[p]){
        if(!used[x] && s[x] == 'B'){
            used[p] = true;
            res |= dfs4(x);
        }
        if(res){
            // cout << x << ' ';
            break;
        }
    }
    return res;
}

bool dfs2(int &p){
    bool res = false;
    for(auto &x : g[p]){
        if(!used[x] && s[x] == 'B'){
            used[p] = true;
            res |= dfs3(x);
        }
        if(res){
            // cout << x << ' ';
            break;
        }
    }
    return res;
}

bool dfs1(int &p){
    bool res = false;
    used2[p] = true;
    for(auto &x : g[p]){
        if(!used[x] && s[x] == 'A'){
            used[p] = true;
            res |= dfs2(x);
        }
        if(res){
            // cout << x << ' ';
            break;
        }
    }
    return res;
}

int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    cin >> s;
    g.resize(n);
    REP(i, m){
        int a, b;
        scanf("%d %d", &a, &b);
        a--;b--;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    
    used.resize(n);
    used2.resize(n);
    vector<int> t;
    REP(i, n){
        if(s[i] == 'A'){
            if(!used2[i]){
                used.assign(n, false);
                goal = i;
                if(dfs1(i)){
                    printf("Yes\n");
                    return 0;
                }
            }
            else{
                t.push_back(i);
            }
        }
    }
    for(int i = 1; i < t.size(); i += 4){
        used.assign(n, false);
        goal = t[i];
        if(dfs1(t[i])){
            printf("Yes\n");
            return 0;
        }
    }
    printf("No\n");
}