#include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

int n,m;
string s;
vector<int> g[200001];
int deg[200001][2];
bool used[200001];
int main(){
    cin>>n>>m>>s;
    int ai,bi;
    rep(i,m){
        cin>>ai>>bi;
        --ai,--bi;
        g[ai].pb(bi);
        g[bi].pb(ai);
        ++deg[ai][s[bi]-'A'];
        ++deg[bi][s[ai]-'A'];
    }
    queue<int> q;
    rep(i,n)if(deg[i][0]==0||deg[i][1]==0) q.push(i);
    int cnt=0;
    while(!q.empty()){
        int v=q.front();q.pop();
        if(!used[v]){
            ++cnt;
            used[v]=true;
            rep(i,g[v].size()){
                --deg[g[v][i]][s[v]-'A'];
                if(deg[g[v][i]][s[v]-'A']==0||deg[g[v][i]][1-(s[v]-'A')]==0) q.push(g[v][i]);
            }
        }
    }
    if(cnt==n) cout<<"No"<<endl;
    else cout<<"Yes"<<endl;
    return 0;
}
### Prompt

Generate a Cpp solution to the following problem:

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`.

Examples

Input

2 3
AB
1 1
1 2
2 2


Output

Yes


Input

4 3
ABAB
1 2
2 3
3 1


Output

No


Input

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


Output

Yes


Input

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


Output

No

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define rep(i,n)   FOR(i,0,n)
#define pb emplace_back
typedef long long ll;
typedef pair<int,int> pint;

int n,m;
string s;
vector<int> g[200001];
int deg[200001][2];
bool used[200001];
int main(){
    cin>>n>>m>>s;
    int ai,bi;
    rep(i,m){
        cin>>ai>>bi;
        --ai,--bi;
        g[ai].pb(bi);
        g[bi].pb(ai);
        ++deg[ai][s[bi]-'A'];
        ++deg[bi][s[ai]-'A'];
    }
    queue<int> q;
    rep(i,n)if(deg[i][0]==0||deg[i][1]==0) q.push(i);
    int cnt=0;
    while(!q.empty()){
        int v=q.front();q.pop();
        if(!used[v]){
            ++cnt;
            used[v]=true;
            rep(i,g[v].size()){
                --deg[g[v][i]][s[v]-'A'];
                if(deg[g[v][i]][s[v]-'A']==0||deg[g[v][i]][1-(s[v]-'A')]==0) q.push(g[v][i]);
            }
        }
    }
    if(cnt==n) cout<<"No"<<endl;
    else cout<<"Yes"<<endl;
    return 0;
}
```