#include <bits/stdc++.h>
//#pragma GCC optimize("O3")
using namespace std;
#define ll long long
#define pb push_back
#define fr first
#define sc second
#define MAX ((ll)(1e17+100))
#define MX ((ll)(1e6+100))
#define ARRS ((ll)(2e6+100))
#define HS ((ll)(233))
#define MOD ((ll)(1e9+7))
#define EP ((double)(1e-9))
#define LG 21
#define mul(a,b) a=((a)*(b))%MOD
using namespace std;
ll n,m;
string s;
vector<ll> v[ARRS];
ll ct[ARRS][2];
ll f[ARRS];
ll t[ARRS];
stack<ll> g;


int main(){
	cin>>n>>m;
	cin>>s;
	for(int i=1; i<=n; i++)
		t[i]=(s[i-1]=='A');
	for(int i=0; i<m; i++){
		ll k,p;
		cin>>k>>p;
		v[k].pb(p);
		v[p].pb(k);
		ct[p][t[k]]++;
		ct[k][t[p]]++;
	}
	for(int i=1; i<=n; i++){
		if(!ct[i][0]||!ct[i][1])g.push(i);
	}
	while(g.size()){
		auto x=g.top();
		g.pop();
		if(f[x])continue;
		f[x]=1;
		for(auto y:v[x]){
			ct[y][t[x]]--;
			if(!ct[y][0]||!ct[y][1])g.push(y);
		}
	}
	for(int i=1; i<=n; i++){
		if(!f[i]){
			cout<<"Yes";
			return 0;
		}
	}
	cout<<"No";
}

### Prompt

Please create a solution in CPP 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 <bits/stdc++.h>
//#pragma GCC optimize("O3")
using namespace std;
#define ll long long
#define pb push_back
#define fr first
#define sc second
#define MAX ((ll)(1e17+100))
#define MX ((ll)(1e6+100))
#define ARRS ((ll)(2e6+100))
#define HS ((ll)(233))
#define MOD ((ll)(1e9+7))
#define EP ((double)(1e-9))
#define LG 21
#define mul(a,b) a=((a)*(b))%MOD
using namespace std;
ll n,m;
string s;
vector<ll> v[ARRS];
ll ct[ARRS][2];
ll f[ARRS];
ll t[ARRS];
stack<ll> g;


int main(){
	cin>>n>>m;
	cin>>s;
	for(int i=1; i<=n; i++)
		t[i]=(s[i-1]=='A');
	for(int i=0; i<m; i++){
		ll k,p;
		cin>>k>>p;
		v[k].pb(p);
		v[p].pb(k);
		ct[p][t[k]]++;
		ct[k][t[p]]++;
	}
	for(int i=1; i<=n; i++){
		if(!ct[i][0]||!ct[i][1])g.push(i);
	}
	while(g.size()){
		auto x=g.top();
		g.pop();
		if(f[x])continue;
		f[x]=1;
		for(auto y:v[x]){
			ct[y][t[x]]--;
			if(!ct[y][0]||!ct[y][1])g.push(y);
		}
	}
	for(int i=1; i<=n; i++){
		if(!f[i]){
			cout<<"Yes";
			return 0;
		}
	}
	cout<<"No";
}

```