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

string s;
multiset<int> edge[200005];
int memo[200005][2];

int main(int argc, char const *argv[])
{
	int n,m;
	cin>>n>>m>>s;
	int a,b;
	for (int i = 0; i < m; ++i)
	{
		cin>>a>>b;
		--a;--b;
		if (edge[a].count(b)>0)
		{
			continue;
		}
		edge[a].insert(b);
		++memo[a][s[b]-'A'];
		if (a!=b)
		{
			edge[b].insert(a);
			++memo[b][s[a]-'A'];
		}
	}
	multiset<int> que;
	for (int i = 0; i < n; ++i)
	{
		if (memo[i][0]==0||memo[i][1]==0)
		{
			que.insert(i);
			while (que.size()>0)
			{
				int now=*que.begin();
				for (auto itr = edge[now].begin(); itr != edge[now].end();)
				{
					int v=*itr;
					itr=edge[now].erase(itr);
					--memo[now][s[v]-'A'];
					edge[v].erase(now);
					--memo[v][s[now]-'A'];
					if (memo[v][0]==0||memo[v][1]==0)
					{
						que.insert(v);
					}
				}
				que.erase(now);
			}
		}
	}
	int f=0;
	for (int i = 0; i < n; ++i)
	{
		if (edge[i].size()>0)
		{
			f=1;
		}
	}
	if (f)
	{
		cout<<"Yes\n";
	}
	else
	{
		cout<<"No\n";
	}
	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;

string s;
multiset<int> edge[200005];
int memo[200005][2];

int main(int argc, char const *argv[])
{
	int n,m;
	cin>>n>>m>>s;
	int a,b;
	for (int i = 0; i < m; ++i)
	{
		cin>>a>>b;
		--a;--b;
		if (edge[a].count(b)>0)
		{
			continue;
		}
		edge[a].insert(b);
		++memo[a][s[b]-'A'];
		if (a!=b)
		{
			edge[b].insert(a);
			++memo[b][s[a]-'A'];
		}
	}
	multiset<int> que;
	for (int i = 0; i < n; ++i)
	{
		if (memo[i][0]==0||memo[i][1]==0)
		{
			que.insert(i);
			while (que.size()>0)
			{
				int now=*que.begin();
				for (auto itr = edge[now].begin(); itr != edge[now].end();)
				{
					int v=*itr;
					itr=edge[now].erase(itr);
					--memo[now][s[v]-'A'];
					edge[v].erase(now);
					--memo[v][s[now]-'A'];
					if (memo[v][0]==0||memo[v][1]==0)
					{
						que.insert(v);
					}
				}
				que.erase(now);
			}
		}
	}
	int f=0;
	for (int i = 0; i < n; ++i)
	{
		if (edge[i].size()>0)
		{
			f=1;
		}
	}
	if (f)
	{
		cout<<"Yes\n";
	}
	else
	{
		cout<<"No\n";
	}
	return 0;
}

```