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.BufferedReader;
import java.util.*;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Code {	
	public static void main(String[] args) {
		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++) {
			int a=sc.nextInt(),b=sc.nextInt();
			x[i]=a;
			y[i]=b;
		}
		
		int answer=solve(n,m,s,x,y);
		System.out.println(answer);
	}	
	
	 static String ss;
	 static ArrayList<Integer>[] ln;
     static boolean[] visited;
     static boolean[] universal;
     static boolean flag;
     static int[][] temp;
     static int ans;
	static int solve(int n,int m,String s,int[] x,int[] y) {
		ss=s;
		 visited = new boolean[n + 1];
         universal = new boolean[n + 1];
         ln = new ArrayList[n + 1];
         for (int i = 1; i <= n; i++) ln[i] = new ArrayList<>();
         flag = false;
         temp = new int[n + 1][26];
         ans = Integer.MIN_VALUE;

         for (int i = 0; i < m; i++) ln[x[i]].add(y[i]);
         for (int i = 1; i <= n; i++) dfs(i);
         if (flag==true) return -1;
         
         return ans;
	}
	
	private static  void dfs(int node) {
        if (universal[node]) return;
        if (visited[node]) return;
        visited[node] = true;
        universal[node] = true;
        Iterator<Integer> it = ln[node].iterator();
        while (it.hasNext()) {
            int temp1 = it.next();
            if (!visited[temp1]) {
                if (!universal[temp1]) dfs(temp1);
                for (int i = 0; i < 26; i++) temp[node][i] = Math.max(temp[temp1][i], temp[node][i]);
            } else {
                flag = true;
                return;
            }
        }
        temp[node][(int) ss.charAt(node - 1) - 97]++;
        for (int i = 0; i < 26; i++) ans = Math.max(ans, temp[node][i]);
        visited[node] = false;
    }
}
 
	
