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.io.*;
import java.util.*;
public class Sol4{
	public static List<Integer> adj[];
	public static int dp[][] = new int[300001][26];
	public static int kk[];
	public static void main(String[] args) throws IOException{
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		adj = new LinkedList[n];
		kk = new int[n];
		String k =sc.next();
		int degree[] = new int[n];
		for(int i=0; i<n; i++) {
			kk[i] = k.charAt(i)-'a';
			adj[i] = new LinkedList<>();
		}
		for(int i=0; i<m; i++) {
			int a = sc.nextInt()-1;
			int b = sc.nextInt()-1;
			adj[a].add(b);
			degree[b]++;
		}
		ArrayDeque<Integer> q = new ArrayDeque<>();
		for(int i=0; i<n; i++) {
			if(degree[i]!=0) continue;
			q.add(i);
		}
		int check =0;
		while(!q.isEmpty()) {
			check++;
			int curr = q.poll();
			dp[curr][kk[curr]]++;
			for(int i:adj[curr]) {
				for(int j=0; j<26; j++) {
					dp[i][j] = Math.max(dp[i][j],dp[curr][j]);
				}
				degree[i]--;
				if(degree[i]==0)q.add(i);
			}
		}
		
		if(check<n)System.out.println(-1);
		else {
			int max = 0;
			for(int i=0;i <n; i++) {
				for(int j=0; j<26; j++) {
					max=Math.max(max, dp[i][j]);
				}
			}
			System.out.println(max);
		}
		sc.close();
	}
}