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.LinkedList;
import java.util.Scanner;

public class Solution {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		int m = sc.nextInt();
		
		char[] nChar = new char[n+1];
		sc.nextLine();
		String s = sc.nextLine();
		for (int i = 1; i <= s.length(); i++) {
			nChar[i] = s.charAt(i-1);
		}
		
		@SuppressWarnings("unchecked")
		LinkedList<Integer>[] G = new LinkedList[n+1];
		for (int i = 0; i <= n; i++) {
			G[i] = new LinkedList<Integer>();
		}
		
		int[] inDeg = new int[n+1];
		
		for (int i = 0; i < m; i++) {
			int u = sc.nextInt();
			int v = sc.nextInt();
			
			G[u].add(v);
			
			inDeg[v]++;
		}
		
		int[][] dp = new int[n+1][26];
		
		for (int i = 1; i <= n; i++) {
			dp[i][nChar[i]-'a'] = 1;
		}
		
		LinkedList<Integer> Q = new LinkedList<Integer>();
		for (int i = 1; i <= n; i++) {
			if (inDeg[i] == 0) {
				Q.addLast(i);
			}
		}
		
		int cnt = 0;
		int max = 1;
		while(!Q.isEmpty()) {
			cnt++;
			
			int u = Q.removeFirst();
			for (int v : G[u]) {
				for (int i = 0; i < 26; i++) {
					dp[v][i] = Math.max(dp[v][i], dp[u][i] + (nChar[v]-'a' == i ? 1 : 0));
					max = Math.max(max, dp[v][i]);
				}
				
				inDeg[v]--;
				
				if (inDeg[v] == 0) {
					Q.addLast(v);
				}
			}
		}
		
		System.out.println(cnt == n ? max : -1);
		sc.close();
	}
}