#include<cstdio>
#include<vector>
using namespace std;

const int MAX_N = 200000;
char s[MAX_N+1];
vector<int> G[MAX_N];
bool valid[MAX_N];
int a_deg[MAX_N];
int b_deg[MAX_N];

void invalidate(int v) {
	valid[v] = false;
	for(int i=0;i<G[v].size();i++) {
		int u = G[v][i];
		if(valid[u]) {
			if(s[v] == 'A') a_deg[u]--;
			else b_deg[u]--;
			if(a_deg[u] == 0 || b_deg[u] == 0) invalidate(u);
		}
	}
}

void add_edge(int u, int v) {
	G[u].push_back(v);
	if(s[v] == 'A') a_deg[u]++;
	else b_deg[u]++;
}

int main() {
	int n, m;
	scanf("%d %d", &n, &m);
	scanf("%s", s);
	for(int i=0;i<m;i++) {
		int a, b;
		scanf("%d %d", &a, &b);
		a--; b--;
		add_edge(a, b);
		add_edge(b, a);
	}
	for(int i=0;i<n;i++) valid[i] = true;
	for(int i=0;i<n;i++) if(valid[i]) {
		if(a_deg[i] == 0 || b_deg[i] == 0) invalidate(i);
	}
	bool empty = true;
	for(int i=0;i<n;i++) if(valid[i]) empty = false;
	printf("%s\n", empty ? "No" : "Yes");
	return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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<cstdio>
#include<vector>
using namespace std;

const int MAX_N = 200000;
char s[MAX_N+1];
vector<int> G[MAX_N];
bool valid[MAX_N];
int a_deg[MAX_N];
int b_deg[MAX_N];

void invalidate(int v) {
	valid[v] = false;
	for(int i=0;i<G[v].size();i++) {
		int u = G[v][i];
		if(valid[u]) {
			if(s[v] == 'A') a_deg[u]--;
			else b_deg[u]--;
			if(a_deg[u] == 0 || b_deg[u] == 0) invalidate(u);
		}
	}
}

void add_edge(int u, int v) {
	G[u].push_back(v);
	if(s[v] == 'A') a_deg[u]++;
	else b_deg[u]++;
}

int main() {
	int n, m;
	scanf("%d %d", &n, &m);
	scanf("%s", s);
	for(int i=0;i<m;i++) {
		int a, b;
		scanf("%d %d", &a, &b);
		a--; b--;
		add_edge(a, b);
		add_edge(b, a);
	}
	for(int i=0;i<n;i++) valid[i] = true;
	for(int i=0;i<n;i++) if(valid[i]) {
		if(a_deg[i] == 0 || b_deg[i] == 0) invalidate(i);
	}
	bool empty = true;
	for(int i=0;i<n;i++) if(valid[i]) empty = false;
	printf("%s\n", empty ? "No" : "Yes");
	return 0;
}

```