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 May {
 

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		 int n = sc.nextInt();
	        int m = sc.nextInt();
	        char[] c = sc.next().toCharArray();
	        int[][] ans = new int[n][26];
	        List<Integer>[] edges = new ArrayList[n];
	        List<Integer>[] ed = new ArrayList[n];
	        for (int i = 0; i < n; i++) {
	            edges[i] = new ArrayList<>();
	            ed[i] = new ArrayList<>();
	        }
	        for (int i = 0; i < m; i++) {
	            int a = sc.nextInt() - 1;
	            int b = sc.nextInt() - 1;
	            edges[a].add(b);
	            ed[b].add(a);
	        }
	        visited = new int[n];
	        for (int i = 0; i < n; i++) {
	            if (visited[i] == 0) {
	                if (dfs(i, edges)) {
	                    System.out.println(-1);
	                    return;
	                }
	            }
	        }
	        for (int i = 0; i < n; i++) {
	            int q = sort.get(i);
	            for (int w : edges[q]) {
	                for (int j = 0; j < 26; j++) {
	                    ans[q][j] = Math.max(ans[w][j], ans[q][j]);
	                }
	            }
	            ans[q][c[q] - 'a']++;
	        }
	        int max = 0;
	        for (int i = 0; i < n; i++) {
	            for (int j = 0; j < 26; j++) {
	                max = Math.max(max, ans[i][j]);
	            }
	        }
	        System.out.println(max);
	    }
	
	    private static int[] visited;
	    private static List<Integer> sort = new ArrayList<>();

	    private static boolean dfs(int n, List<Integer>[] edges) {
	        visited[n] = 1;
	        for (int q : edges[n]) {
	            if (visited[q] == 1) {
	                return true;
	            }
	            if (visited[q] == 0) {
	                if (dfs(q, edges)) {
	                    return true;
	                }
	            }
	        }
	        visited[n] = 2;
	        sort.add(n);
	        return false;
	    }

	
}
