#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
const int N=200050;
const int M=2*N;
char col[N];
vector<int> E[M];
int cyc[M];
bool cycle=0;
void DFS(int u)
{
	cyc[u]=1;
	for(int v:E[u])
	{
		if(!cyc[v]) DFS(v);
		else if(cyc[v]==1) cycle=1;
	}
	cyc[u]=2;
}
int main()
{
	int n,m,i,u,v;
	scanf("%i %i",&n,&m);
	scanf("%s",col+1);
	for(i=1;i<=m;i++)
	{
		scanf("%i %i",&u,&v);
		if(col[u]==col[v]) E[u].pb(v+n),E[v].pb(u+n);
		else E[u+n].pb(v),E[v+n].pb(u);
	}
	for(i=1;i<=n;i++) if(!cyc[i]) DFS(i);
	if(cycle) printf("Yes\n");
	else printf("No\n");
	return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 pb push_back
#define mp make_pair
const int N=200050;
const int M=2*N;
char col[N];
vector<int> E[M];
int cyc[M];
bool cycle=0;
void DFS(int u)
{
	cyc[u]=1;
	for(int v:E[u])
	{
		if(!cyc[v]) DFS(v);
		else if(cyc[v]==1) cycle=1;
	}
	cyc[u]=2;
}
int main()
{
	int n,m,i,u,v;
	scanf("%i %i",&n,&m);
	scanf("%s",col+1);
	for(i=1;i<=m;i++)
	{
		scanf("%i %i",&u,&v);
		if(col[u]==col[v]) E[u].pb(v+n),E[v].pb(u+n);
		else E[u+n].pb(v),E[v+n].pb(u);
	}
	for(i=1;i<=n;i++) if(!cyc[i]) DFS(i);
	if(cycle) printf("Yes\n");
	else printf("No\n");
	return 0;
}
```