#include <bits/stdc++.h>
using namespace std;

#define N 100000*2
#define INF 1LL << 60

int n, m;
string s;
vector<int> g[N];

int numA[N];
int numB[N];
bool deleted[N];

void del( int i )
{
	if( deleted[i] ) {
		return;
	}
	deleted[i] = true;
	for( auto j : g[i] ) {
		if( deleted[j] ) {
			continue;
		}
		if( s[i] == 'A' ) {
			numA[j]--;
		}
		else {
			numB[j]--;
		}
		if( numA[j] == 0 || numB[j] == 0 ) {
			del( j );
		}
	}
}

int main()
{
	int a, b;

	cin >> n >> m;
	cin >> s;
	for( int i = 0; i < m; i++ ) {
		cin >> a >> b;
		a--;
		b--;
		g[a].push_back( b );
		g[b].push_back( a );
		if( s[b] == 'A' ) {
			numA[a]++;
		}
		else {
			numB[a]++;
		}
		if( s[a] == 'A' ) {
			numA[b]++;
		}
		else {
			numB[b]++;
		}
	}

	for( int i = 0; i < n; i++ ) {
		if( numA[i] == 0 || numB[i] == 0 ) {
			del( i );
		}
	}
	int flag = 0;
	for( int i = 0; i < n; i++ ) {
		if( !deleted[i] ) {
			flag = 1;
		}
	}

	if( flag ) {
		cout << "Yes" << endl;
	}
	else {
		cout << "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;

#define N 100000*2
#define INF 1LL << 60

int n, m;
string s;
vector<int> g[N];

int numA[N];
int numB[N];
bool deleted[N];

void del( int i )
{
	if( deleted[i] ) {
		return;
	}
	deleted[i] = true;
	for( auto j : g[i] ) {
		if( deleted[j] ) {
			continue;
		}
		if( s[i] == 'A' ) {
			numA[j]--;
		}
		else {
			numB[j]--;
		}
		if( numA[j] == 0 || numB[j] == 0 ) {
			del( j );
		}
	}
}

int main()
{
	int a, b;

	cin >> n >> m;
	cin >> s;
	for( int i = 0; i < m; i++ ) {
		cin >> a >> b;
		a--;
		b--;
		g[a].push_back( b );
		g[b].push_back( a );
		if( s[b] == 'A' ) {
			numA[a]++;
		}
		else {
			numB[a]++;
		}
		if( s[a] == 'A' ) {
			numA[b]++;
		}
		else {
			numB[b]++;
		}
	}

	for( int i = 0; i < n; i++ ) {
		if( numA[i] == 0 || numB[i] == 0 ) {
			del( i );
		}
	}
	int flag = 0;
	for( int i = 0; i < n; i++ ) {
		if( !deleted[i] ) {
			flag = 1;
		}
	}

	if( flag ) {
		cout << "Yes" << endl;
	}
	else {
		cout << "No" << endl;
	}

	return 0;
}

```