#include <bits/stdc++.h>
#define maxn 200100

using namespace std;

int n,m,in[maxn],num[maxn][3],col[maxn];
string s;
vector<int> g[maxn],dead;

int main() {
   // freopen("in.txt","r",stdin);
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin>>n>>m;
    cin>>s;
    for (int i=0;i<n;i++) col[i]=(s[i]=='A');
    for (int i=1,u,v;i<=m;i++) {
        cin>>u>>v;
        u--; v--;
        g[u].push_back(v);
        g[v].push_back(u);
        num[u][col[v]]++;
        num[v][col[u]]++;
    }
    for (int i=0;i<n;i++)
        if (num[i][0]==0 || num[i][1]==0) {
            dead.push_back(i);
            in[i]=1;
        }
    for (int i=0;i<dead.size();i++) {
        int u = dead[i];
        for (auto v : g[u]) {
            if (in[v]) continue;
            num[v][col[u]]--;
            if (num[v][0]==0 || num[v][1]==0) {
                dead.push_back(v);
                in[v]=1;
            }
        }
    }
    if (dead.size()==n) cout<<"No";
    else cout<<"Yes";
}

### Prompt

Your challenge is to write 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>
#define maxn 200100

using namespace std;

int n,m,in[maxn],num[maxn][3],col[maxn];
string s;
vector<int> g[maxn],dead;

int main() {
   // freopen("in.txt","r",stdin);
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin>>n>>m;
    cin>>s;
    for (int i=0;i<n;i++) col[i]=(s[i]=='A');
    for (int i=1,u,v;i<=m;i++) {
        cin>>u>>v;
        u--; v--;
        g[u].push_back(v);
        g[v].push_back(u);
        num[u][col[v]]++;
        num[v][col[u]]++;
    }
    for (int i=0;i<n;i++)
        if (num[i][0]==0 || num[i][1]==0) {
            dead.push_back(i);
            in[i]=1;
        }
    for (int i=0;i<dead.size();i++) {
        int u = dead[i];
        for (auto v : g[u]) {
            if (in[v]) continue;
            num[v][col[u]]--;
            if (num[v][0]==0 || num[v][1]==0) {
                dead.push_back(v);
                in[v]=1;
            }
        }
    }
    if (dead.size()==n) cout<<"No";
    else cout<<"Yes";
}

```