#include <iostream>
#include <queue>
#include <cstring>
#include <vector>
using namespace std;
int t[200005][2];
vector<int> g[200005];
bool used[200005];
queue<int> v;
int main() {
  int n,m;
  string s;
  cin >> n >> m >> s;
  for(int i = 0; i < m; i++) {
    int a,b;
    cin >> a >> b;
    a--,b--;
    g[a].push_back(b);
    g[b].push_back(a);
    if(s[a] == 'A') t[b][0]++;
    if(s[a] == 'B') t[b][1]++;
    if(s[b] == 'A') t[a][0]++;
    if(s[b] == 'B') t[a][1]++;
  }
  for(int i = 0; i < n; i++) {
    if(t[i][0] == 0 || t[i][1] == 0) {
      v.push(i);
      used[i] = true;
    }
  }
  int ans = v.size();
  // cout << ans << endl;
  while(!v.empty()) {
    int u = v.front();
    v.pop();
    int tmp = s[u] - 'A';
    for(int i = 0; i < g[u].size(); i++) {
      int e = g[u][i];
      t[e][tmp]--;
      if((t[e][0] == 0 || t[e][1] == 0) && !used[e]) {
	v.push(e);
	ans++;
	used[e] = true;
      }
    }
  }
  // cout << ans << endl;
  if(ans == n) cout << "No" << endl;
  else cout << "Yes" << endl;
}

  
  

### 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 <iostream>
#include <queue>
#include <cstring>
#include <vector>
using namespace std;
int t[200005][2];
vector<int> g[200005];
bool used[200005];
queue<int> v;
int main() {
  int n,m;
  string s;
  cin >> n >> m >> s;
  for(int i = 0; i < m; i++) {
    int a,b;
    cin >> a >> b;
    a--,b--;
    g[a].push_back(b);
    g[b].push_back(a);
    if(s[a] == 'A') t[b][0]++;
    if(s[a] == 'B') t[b][1]++;
    if(s[b] == 'A') t[a][0]++;
    if(s[b] == 'B') t[a][1]++;
  }
  for(int i = 0; i < n; i++) {
    if(t[i][0] == 0 || t[i][1] == 0) {
      v.push(i);
      used[i] = true;
    }
  }
  int ans = v.size();
  // cout << ans << endl;
  while(!v.empty()) {
    int u = v.front();
    v.pop();
    int tmp = s[u] - 'A';
    for(int i = 0; i < g[u].size(); i++) {
      int e = g[u][i];
      t[e][tmp]--;
      if((t[e][0] == 0 || t[e][1] == 0) && !used[e]) {
	v.push(e);
	ans++;
	used[e] = true;
      }
    }
  }
  // cout << ans << endl;
  if(ans == n) cout << "No" << endl;
  else cout << "Yes" << endl;
}

  
  

```