#include <iostream>
#include <vector>

using namespace std;

int n, m;
string s;
vector<int> v[202020];
int d1[202020], d2[202020];
int z[202020];

void dfs(int x) {
	if (d1[x] >= 1 && d2[x] >= 1) return;
	if (z[x]) return;
	z[x] = 1;
	for (auto u : v[x]) {
		if (s[x] == 'A') d1[u]--;
		if (s[x] == 'B') d2[u]--;
		if (!z[u] && (d1[u] < 1 || d2[u] < 1)) dfs(u);
	}
}

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cin >> n >> m;
	cin >> s;
	s = "#"+s;
	for (int i = 1; i <= m; i++) {
		int a, b;
		cin >> a >> b;
		v[a].push_back(b);
		v[b].push_back(a);
		if (s[a] == 'A') d1[b]++; else d2[b]++;
		if (s[b] == 'A') d1[a]++; else d2[a]++;
	}
	for (int i = 1; i <= n; i++) {
		dfs(i);
	}
	for (int i = 1; i <= n; i++) {
		if (d1[i] >= 1 && d2[i] >= 1) {
			cout << "Yes\n";
			return 0;
		}
	}
	cout << "No\n";
}

### Prompt

Generate a CPP solution 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 <iostream>
#include <vector>

using namespace std;

int n, m;
string s;
vector<int> v[202020];
int d1[202020], d2[202020];
int z[202020];

void dfs(int x) {
	if (d1[x] >= 1 && d2[x] >= 1) return;
	if (z[x]) return;
	z[x] = 1;
	for (auto u : v[x]) {
		if (s[x] == 'A') d1[u]--;
		if (s[x] == 'B') d2[u]--;
		if (!z[u] && (d1[u] < 1 || d2[u] < 1)) dfs(u);
	}
}

int main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cin >> n >> m;
	cin >> s;
	s = "#"+s;
	for (int i = 1; i <= m; i++) {
		int a, b;
		cin >> a >> b;
		v[a].push_back(b);
		v[b].push_back(a);
		if (s[a] == 'A') d1[b]++; else d2[b]++;
		if (s[b] == 'A') d1[a]++; else d2[a]++;
	}
	for (int i = 1; i <= n; i++) {
		dfs(i);
	}
	for (int i = 1; i <= n; i++) {
		if (d1[i] >= 1 && d2[i] >= 1) {
			cout << "Yes\n";
			return 0;
		}
	}
	cout << "No\n";
}

```