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.*;
 
public class D {
 
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt(), m = sc.nextInt();
		String s = sc.next();
		
		ArrayList<Integer>[] out = new ArrayList[n+1]; 
		
		for(int i=1;i<=n;i++)
			out[i] = new ArrayList<Integer>();
		
		int deg[] = new int[n+1];
		
		while(m-->0){
			int a = sc.nextInt();
			int b = sc.nextInt();
			deg[b]++;
			out[a].add(b);
		}
 
		Queue<Integer> q = new ArrayDeque<>();
		int f[][] = new int[n+1][26];
		
		for(int i=1;i<=n;i++){
			if(deg[i] == 0){
				q.offer(i);
				f[i][s.charAt(i-1)-'a']++;
			}
		}
		
		int count = 0;
		while(count < n && q.size()>0){
			int v = q.poll();
			
			for(int i : out[v]){
				
				for(int j=0;j<26;j++){
					if(j==s.charAt(i-1)-'a')
						f[i][j] = Math.max(f[v][j]+1, f[i][j]);
					else
						f[i][j] = Math.max(f[v][j], f[i][j]);
				}
				
				if(--deg[i] == 0){
					q.offer(i);
				}
			}
			
			count++;
		}
		
		if(count<n){
			System.out.println(-1);
		}
		else{
			int ans = 0;
			for(int arr[] : f){
				for(int i : arr){
					ans = Math.max(i, ans);
				}
			}
			
			System.out.println(ans);
		}
	}
	
}