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 class Codechef
{
	static int[][] memo;
	static int[] vis;
	static ArrayList<Integer> a[];
	static int[] letter;
	public static void main (String[] args)
	{
		Scanner sc=new Scanner(System.in);
		int n=sc.nextInt();
		int m=sc.nextInt();
		String s=sc.next();
		a=new ArrayList[n+1];
		for(int i=0;i<=n;i++) {
			a[i]=new ArrayList<Integer>();
		}
		for(int i=0;i<m;i++) {
			int a1=sc.nextInt();
			int a2=sc.nextInt();
			a[a1].add(a2);
		}
		letter=new int[n+1];
		for(int i=1;i<=n;i++) {
			letter[i]=s.charAt(i-1)-'a';
		}
		vis=new int[n+1];
		memo=new int[n+1][26];
		for(int i=0;i<=n;i++) {
			if(vis[i]==0 && dfs(i)) {
				System.out.println(-1);
				return;
			}
		}
		int max=0;
		for(int i=0;i<=n;i++) {
			for(int j=0;j<26;j++) {
				max=Math.max(memo[i][j],max);
			}
		}
		System.out.println(max);
	}
	public static boolean dfs(int v) {
		vis[v]=1;
		for(int neigh : a[v]) {
			if(vis[neigh]==1)
				return true;
			if(vis[neigh]==0 && dfs(neigh))
				return true;
			for(int i=0;i<26;i++)
				memo[v][i]=Math.max(memo[neigh][i], memo[v][i]);
		}
		memo[v][letter[v]]++;
		vis[v]=2;
		return false;
	}
}
