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.*;
import java.io.*;

public class D {

	public static class node {
		
		LinkedList<node> adj;
		int chr;
		int[] arr;
		boolean calc;
		int id;
		int inc;
		
		public node(char a, int id) {
			this.id = id;
			adj = new LinkedList<node>();
			chr = a - 'a';
			arr = new int[26];
			calc = false;
			inc = 0;
		}
		
		public int count() {
			if(!calc) this.calcu();
			return arr[chr];
		}
		
		public void calcu() {
			for(node n : adj) {
				if(!n.calc) n.calcu();
				for(int i = 0; i < 26; i++) {
					if(n.arr[i] > this.arr[i]) this.arr[i] = n.arr[i];
				}
			}
			this.arr[chr]++;
			calc = true;
		}
	}

	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		String[] ree = in.readLine().split("\\s++");
		int n = Integer.parseInt(ree[0]);
		int m = Integer.parseInt(ree[1]);
		char[] r = in.readLine().toCharArray();
		node[] nodes = new node[n];
		for(int i = 0; i < n; i++) {
			nodes[i] = new node(r[i], i);
		}
		for(int i = 0; i < m; i++) {
			ree = in.readLine().split("\\s++");
			int from = Integer.parseInt(ree[0]);
			int to = Integer.parseInt(ree[1]);
			nodes[from - 1].adj.offer(nodes[to - 1]);
			nodes[to - 1].inc++;
		}
		boolean loop = false;
		LinkedList<node> q = new LinkedList<node>();
		for(int i = 0; i < n; i++) {
			if(nodes[i].inc == 0) q.offer(nodes[i]);
		}
		while(!q.isEmpty()) {
			node c = q.poll();
			for(node ee : c.adj) {
				ee.inc--;
				if(ee.inc == 0) {
					q.offer(ee);
				}
			}
		}
		for(int i = 0; i < n; i++) {
			if(nodes[i].inc > 0) loop = true;
		}
		if(loop) {
			System.out.println(-1);
			return;
		}
		int best = 1;
		for(int i = 0; i < n; i++) {
			if(nodes[i].count() > best) best = nodes[i].count();
		}
		System.out.println(best);
	}
}