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;
#define int long long
#define REP(i,m,n) for(int i=(m);i<(n);i++)
#define rep(i,n) REP(i,0,n)
#define pb push_back
#define all(a) a.begin(),a.end()
#define rall(c) (c).rbegin(),(c).rend()
#define mp make_pair
#define endl '\n'
#define fi first
#define se second
typedef long long ll;
typedef pair<ll,ll> pll;
typedef long double ld;
const ll inf=1e9+7;
const ll mod=998244353;
signed main(){
    ll n,m;cin>>n>>m;
    string s;cin>>s;
    vector<vector<ll> >G(n);
    rep(i,m){
        ll a,b;cin>>a>>b;
        a--;b--;
        G[a].pb(b);
        G[b].pb(a);
    }
    rep(i,n){
        sort(all(G[i]));
        REP(j,1,G[i].size()){
            if(G[i][j]==G[i][j-1]){
                G[i].erase(G[i].begin()+j);
            }
        }
    }
    vector<ll>a(n),b(n);
    queue<ll>q;
    rep(i,n){
        for(auto e:G[i]){
            if(s[e]=='A')a[i]++;
            if(s[e]=='B')b[i]++;
        }
        if(a[i]>0&&b[i]>0)continue;
        q.push(i);
    }
    vector<ll>used(n);
    while(!q.empty()){
        ll k=q.front();
        q.pop();
        used[k]=1;
        for(auto e:G[k]){
            if(used[e])continue;
            if(e==k)continue;
            if(s[k]=='A')a[e]--;
            if(s[k]=='B')b[e]--;
            if(a[e]>0&&b[e]>0)continue;
            q.push(e);
        }
    }
    ll cnt=0;
    rep(i,n){
        if(used[i]==0)cnt++;
    }
    cout<<(cnt>0?"Yes":"No")<<endl;
}