#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll ans = 0;

int main()
{
ll n , m ;
cin >> n >> m ;
string s ;
cin >> s ;
vector<vector<ll>> g(n);
vector<vector<ll>> num(n,vector<ll>(2));
rep(i, m)
{
  ll a, b;
  cin >> a >> b;
  g[a - 1].push_back(b - 1);
  g[b - 1].push_back(a - 1); //無向辺
  if(s[b-1]=='A')num[a-1][0]++;
  if(s[b-1]=='B')num[a-1][1]++;
  if(s[a-1]=='A')num[b-1][0]++;
  if(s[a-1]=='B')num[b-1][1]++;
}
vector<bool> liv(n,1);
queue<ll> q;
rep(i,n)if(num[i][0]==0||num[i][1]==0){q.push(i);liv[i]=0;}
while(!q.empty()){
  ll p=q.front();
  q.pop();
  ll a=((s[p]=='A') ? 0 : 1 );
  for(auto r:g[p]){
    if(liv[r]==0)continue;
    num[r][a]--;
    if(num[r][a]==0){q.push(r);liv[r]=0;}
  }
}
bool ok=0;
rep(i,n)if(num[i][0]>0&&num[i][1]>0)ok=1;
cout << (ok ? "Yes" : "No" ) << endl;
return 0;
}
### Prompt

In cpp, your task is to solve 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;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll ans = 0;

int main()
{
ll n , m ;
cin >> n >> m ;
string s ;
cin >> s ;
vector<vector<ll>> g(n);
vector<vector<ll>> num(n,vector<ll>(2));
rep(i, m)
{
  ll a, b;
  cin >> a >> b;
  g[a - 1].push_back(b - 1);
  g[b - 1].push_back(a - 1); //無向辺
  if(s[b-1]=='A')num[a-1][0]++;
  if(s[b-1]=='B')num[a-1][1]++;
  if(s[a-1]=='A')num[b-1][0]++;
  if(s[a-1]=='B')num[b-1][1]++;
}
vector<bool> liv(n,1);
queue<ll> q;
rep(i,n)if(num[i][0]==0||num[i][1]==0){q.push(i);liv[i]=0;}
while(!q.empty()){
  ll p=q.front();
  q.pop();
  ll a=((s[p]=='A') ? 0 : 1 );
  for(auto r:g[p]){
    if(liv[r]==0)continue;
    num[r][a]--;
    if(num[r][a]==0){q.push(r);liv[r]=0;}
  }
}
bool ok=0;
rep(i,n)if(num[i][0]>0&&num[i][1]>0)ok=1;
cout << (ok ? "Yes" : "No" ) << endl;
return 0;
}
```