You are given a graph with n nodes and m directed edges. One lowercase letter is assigned to each node. We define a path's value as the number of the most frequently occurring letter. For example, if letters on a path are "abaca", then the value of that path is 3. Your task is find a path whose value is the largest.

Input

The first line contains two positive integers n, m (1 ≤ n, m ≤ 300 000), denoting that the graph has n nodes and m directed edges.

The second line contains a string s with only lowercase English letters. The i-th character is the letter assigned to the i-th node.

Then m lines follow. Each line contains two integers x, y (1 ≤ x, y ≤ n), describing a directed edge from x to y. Note that x can be equal to y and there can be multiple edges between x and y. Also the graph can be not connected.

Output

Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead.

Examples

Input

5 4
abaca
1 2
1 3
3 4
4 5


Output

3


Input

6 6
xzyabc
1 2
3 1
2 3
5 4
4 3
6 4


Output

-1


Input

10 14
xzyzyzyzqx
1 2
2 4
3 5
4 5
2 6
6 8
6 5
2 10
3 9
10 9
4 6
1 10
2 8
3 7


Output

4

Note

In the first sample, the path with largest value is 1 → 3 → 4 → 5. The value is 3 because the letter 'a' appears 3 times.
import java.util.*;
import java.lang.*;
import java.io.*;
 
 
public final class Solution
{
	int dp[][] =  new int[300003][30];
	int g[] = new int[300003];
	int p[] = new int[300003];
	int l[]= new int[300003];
	int used[] = new int[300003];
	int lg=0;
	public void insert(int u,int v)
	{
	//	System.out.println("insert");
		lg++;
		g[lg]=v;
		p[lg]=l[u];
		l[u]=lg;
	}
	public void dfs(int u,int k,String a)
	{
		//System.out.println("enter");
		if(used[u]!=0 && used[u]<k)
		{
			System.out.println("-1");
			System.exit(0);
		}
		if(used[u]!=0)
		return;
		used[u]=k;
		int v;
		for(int i=l[u];i!=0;i=p[i])
		{
			v=g[i];
			dfs(v,k+1,a);
			for(int j=0;j<26;j++)
			{
				if(dp[u][j]<dp[v][j])
				dp[u][j]=dp[v][j];
			}
		}
		dp[u][a.charAt(u-1)-'a']++;
		used[u]=1000000;
	}
	public int beauty(int n,int m, String s,int x[],int y[])
	{
	//	System.out.println(n + " " + m + " "+ s);
		for(int[] r : dp){
			Arrays.fill(r,0);
		}
		Arrays.fill(g,0);
		Arrays.fill(p,0);
		Arrays.fill(l,0);
		Arrays.fill(used,0);
		for(int i=0;i<m;i++){
			insert(x[i],y[i]);
		}
		for(int i=1;i<=n;i++){
			dfs(i,1,s);
		}
		int res=0;
		for(int i=1;i<=n;i++)
		{
			for(int j=0;j<26;j++)
			{
				if(dp[i][j]>res)
				res=dp[i][j];
			}
		}
		return res;
	}
	public static void main (String[] args) throws java.lang.Exception
	{
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		String s = sc.next();
		int x[] = new int[m];
		int y[]= new int[m];
		for(int i=0;i<m;i++)
		{
		x[i]=sc.nextInt();
		y[i] = sc.nextInt();
		}
		Solution sol = new Solution();
		int ans = sol.beauty(n,m,s,x,y);
		System.out.println(ans);
	}
}