#include <bits/stdc++.h>

using namespace std;

int n, m, y[200005][2], z[2];
char x[200005];
bool g[200005];
multiset<int> v[200005];
int ta, tb;

void go(int a) {
	g[a] = 1;
	z[x[a]]--;
	while (!v[a].empty()) {
		int tt = *v[a].begin();
		v[a].erase(v[a].begin());
		if (!g[tt]) {
			v[tt].erase(v[tt].find(a));
			y[tt][x[a]]--;
			if (y[tt][0] == 0 || y[tt][1] == 0)
				go(tt);
		}
	}
}

int main() {
	scanf("%d%d%s", &n, &m, x + 1);
	for (int i = 1; i <= n; i++) {
		x[i] -= 'A';
		z[x[i]]++;
	}
	while (m--) {
		scanf("%d%d", &ta, &tb);
		v[ta].insert(tb);
		v[tb].insert(ta);
		y[ta][x[tb]]++;
		y[tb][x[ta]]++;
	}
	for (int i = 1; i <= n; i++)
		if (y[i][0] == 0 || y[i][1] == 0)
			if (!g[i])
				go(i);
	printf(z[0] ? "Yes\n" : "No\n");
	return 0;
}

### 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>

using namespace std;

int n, m, y[200005][2], z[2];
char x[200005];
bool g[200005];
multiset<int> v[200005];
int ta, tb;

void go(int a) {
	g[a] = 1;
	z[x[a]]--;
	while (!v[a].empty()) {
		int tt = *v[a].begin();
		v[a].erase(v[a].begin());
		if (!g[tt]) {
			v[tt].erase(v[tt].find(a));
			y[tt][x[a]]--;
			if (y[tt][0] == 0 || y[tt][1] == 0)
				go(tt);
		}
	}
}

int main() {
	scanf("%d%d%s", &n, &m, x + 1);
	for (int i = 1; i <= n; i++) {
		x[i] -= 'A';
		z[x[i]]++;
	}
	while (m--) {
		scanf("%d%d", &ta, &tb);
		v[ta].insert(tb);
		v[tb].insert(ta);
		y[ta][x[tb]]++;
		y[tb][x[ta]]++;
	}
	for (int i = 1; i <= n; i++)
		if (y[i][0] == 0 || y[i][1] == 0)
			if (!g[i])
				go(i);
	printf(z[0] ? "Yes\n" : "No\n");
	return 0;
}

```