#include <iostream>
#include <vector>
#include <set>
using namespace std;
char s[200005];
vector<int> v[200005];
int cnt[200005][2];
struct cmp
{
	bool operator()(const int &a,const int &b) const
	{
		int ca=min(cnt[a][0],cnt[a][1]),cb=min(cnt[b][0],cnt[b][1]);
		if (ca==cb)
		return (a<b);
		return (ca<cb);
	}
};
set<int,cmp> st;
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	getchar();
	scanf("%s",s);
	while (m--)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		a--;
		b--;
		v[a].push_back(b);
		v[b].push_back(a);
		cnt[a][s[b]-'A']++;
		cnt[b][s[a]-'A']++;
	}
	for (int i=0;i<n;i++)
	st.insert(i);
	while (!st.empty())
	{
		int node=*st.begin();
		if (min(cnt[node][0],cnt[node][1]))
		break;
		st.erase(node);
		for (int u:v[node])
		{
			if (st.find(u)==st.end())
			continue;
			st.erase(u);
			cnt[u][s[node]-'A']--;
			st.insert(u);
		}
	}
	printf(st.empty()? "No":"Yes");
}
### 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 <iostream>
#include <vector>
#include <set>
using namespace std;
char s[200005];
vector<int> v[200005];
int cnt[200005][2];
struct cmp
{
	bool operator()(const int &a,const int &b) const
	{
		int ca=min(cnt[a][0],cnt[a][1]),cb=min(cnt[b][0],cnt[b][1]);
		if (ca==cb)
		return (a<b);
		return (ca<cb);
	}
};
set<int,cmp> st;
int main()
{
	int n,m;
	scanf("%d%d",&n,&m);
	getchar();
	scanf("%s",s);
	while (m--)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		a--;
		b--;
		v[a].push_back(b);
		v[b].push_back(a);
		cnt[a][s[b]-'A']++;
		cnt[b][s[a]-'A']++;
	}
	for (int i=0;i<n;i++)
	st.insert(i);
	while (!st.empty())
	{
		int node=*st.begin();
		if (min(cnt[node][0],cnt[node][1]))
		break;
		st.erase(node);
		for (int u:v[node])
		{
			if (st.find(u)==st.end())
			continue;
			st.erase(u);
			cnt[u][s[node]-'A']--;
			st.insert(u);
		}
	}
	printf(st.empty()? "No":"Yes");
}
```