#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#define pb push_back
#define eb emplace_back
#define F first
#define S second
#define MP make_pair
#define MEM(i,j) memset(i,j,sizeof i)
#define Tie ios::sync_with_stdio(0),cin.tie(0)
#define MOD 1000000007
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

int n,m,a,b,toa[200001],tob[200001],tmp,cnt;
string s;
vector<int> v[200001];
queue<int> q;
bitset<200001> u;

int main()
{
	cin>>n>>m>>s;
	for(int i=0;i<m;i++)
	{
		cin>>a>>b,a--,b--;
		v[a].push_back(b),v[b].push_back(a);
		if(s[a]=='A')
			toa[b]++;
		else tob[b]++;
		if(s[b]=='A')
			toa[a]++;
		else tob[a]++;
	}
	for(int i=0;i<n;i++)
		if(toa[i]==0||tob[i]==0)
			q.push(i),cnt++,u[i]=1;
	while(!q.empty())
	{
		tmp=q.front(),q.pop();
		for(int i:v[tmp])
		{
			if(s[tmp]=='A')
				toa[i]--;
			else
				tob[i]--;
			if(!u[i]&&(toa[i]==0||tob[i]==0))
				q.push(i),cnt++,u[i]=1;
		}
	}
	if(cnt==n)cout << "No\n";
	else cout << "Yes\n";
}
### 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
#pragma GCC optimize("Ofast")
#include <bits/stdc++.h>
#define pb push_back
#define eb emplace_back
#define F first
#define S second
#define MP make_pair
#define MEM(i,j) memset(i,j,sizeof i)
#define Tie ios::sync_with_stdio(0),cin.tie(0)
#define MOD 1000000007
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

int n,m,a,b,toa[200001],tob[200001],tmp,cnt;
string s;
vector<int> v[200001];
queue<int> q;
bitset<200001> u;

int main()
{
	cin>>n>>m>>s;
	for(int i=0;i<m;i++)
	{
		cin>>a>>b,a--,b--;
		v[a].push_back(b),v[b].push_back(a);
		if(s[a]=='A')
			toa[b]++;
		else tob[b]++;
		if(s[b]=='A')
			toa[a]++;
		else tob[a]++;
	}
	for(int i=0;i<n;i++)
		if(toa[i]==0||tob[i]==0)
			q.push(i),cnt++,u[i]=1;
	while(!q.empty())
	{
		tmp=q.front(),q.pop();
		for(int i:v[tmp])
		{
			if(s[tmp]=='A')
				toa[i]--;
			else
				tob[i]--;
			if(!u[i]&&(toa[i]==0||tob[i]==0))
				q.push(i),cnt++,u[i]=1;
		}
	}
	if(cnt==n)cout << "No\n";
	else cout << "Yes\n";
}
```