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

typedef pair<int, int> pii;
const int N = 2e5 + 100;

int n, m, u, v, d[N][2];
bool done[N];
string s;
queue<int> q;
vector<pii> nei[N];

int main() {
	ios::sync_with_stdio(false), cin.tie(0);
	cin >> n >> m >> s;
	while(m--) {
		cin >> u >> v;
		int c = (s[--u] == s[--v]);
		nei[u].push_back(pii(v, c));
		nei[v].push_back(pii(u, c));
	}
	for (int i = 0; i < n; i++)
		for (pii p: nei[i])
			d[i][p.second]++;
	for (int i = 0; i < n; i++)
		if(d[i][0] == 0 || d[i][1] == 0) {
			q.push(i);
			done[i] = true;
		}
	while(q.size()) {
		n--;
		v = q.front();
		q.pop();
		done[v] = true;
		for (pii p: nei[v]) {
			d[p.first][p.second]--;
			if(done[p.first] == false && (d[p.first][0] == 0 || d[p.first][1] == 0)) {
				done[p.first] = true;
				q.push(p.first);
			}
		}
	}
	cout << (n? "Yes": "No");
	return 0;
}
### 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 <bits/stdc++.h>
using namespace std;

typedef pair<int, int> pii;
const int N = 2e5 + 100;

int n, m, u, v, d[N][2];
bool done[N];
string s;
queue<int> q;
vector<pii> nei[N];

int main() {
	ios::sync_with_stdio(false), cin.tie(0);
	cin >> n >> m >> s;
	while(m--) {
		cin >> u >> v;
		int c = (s[--u] == s[--v]);
		nei[u].push_back(pii(v, c));
		nei[v].push_back(pii(u, c));
	}
	for (int i = 0; i < n; i++)
		for (pii p: nei[i])
			d[i][p.second]++;
	for (int i = 0; i < n; i++)
		if(d[i][0] == 0 || d[i][1] == 0) {
			q.push(i);
			done[i] = true;
		}
	while(q.size()) {
		n--;
		v = q.front();
		q.pop();
		done[v] = true;
		for (pii p: nei[v]) {
			d[p.first][p.second]--;
			if(done[p.first] == false && (d[p.first][0] == 0 || d[p.first][1] == 0)) {
				done[p.first] = true;
				q.push(p.first);
			}
		}
	}
	cout << (n? "Yes": "No");
	return 0;
}
```