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.
//package com.company;

import java.io.*;
import java.util.*;

public class Main {
    static long TIME_START, TIME_END;
    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
//        Scanner sc = new Scanner(new FileInputStream("Test.in"));
        PrintWriter pw = new PrintWriter(System.out);
//        PrintWriter pw = new PrintWriter(new FileOutputStream("Test.out"));
//        PrintWriter pw = new PrintWriter(new FileOutputStream("Test.in"));

        Runtime runtime = Runtime.getRuntime();
        long usedMemoryBefore = runtime.totalMemory() - runtime.freeMemory();
        TIME_START = System.currentTimeMillis();
        Task t = new Task();
        t.solve(sc, pw);
        TIME_END = System.currentTimeMillis();
        long usedMemoryAfter = runtime.totalMemory() - runtime.freeMemory();
        pw.close();
//        System.out.println("Memory increased:" + (usedMemoryAfter-usedMemoryBefore) / 1000000 );
//        System.out.println("Time used: " + (TIME_END - TIME_START) + ".");


    }

    //http://codeforces.com/contest/641/problem/D
    //Little Artem and Random Variable
    public static class Task {
        int n, m;
        public class Node {
            Set<Integer> edge;
            int[] val;
            int v;
            public Node(int v) {
                this.edge = new TreeSet<>();
                this.val = new int[26];
                this.v = v;
            }
        }
        Node[] nodes;

        public void solve(Scanner sc, PrintWriter pw) throws IOException {
            n = sc.nextInt();
            m = sc.nextInt();
            String str = sc.next();
            nodes = new Node[n];
            int[] degree = new int[n];
            for (int i = 0; i < n; i++) {
                nodes[i] = new Node(str.charAt(i) - 'a');
            }
            for (int i = 0; i < m; i++) {
                int a = sc.nextInt() - 1;
                int b = sc.nextInt() - 1;
                if (a == b) {
                    pw.println(-1);
                    return;
                }
                if (!nodes[a].edge.contains(b)) {
                    nodes[a].edge.add(b);
                    degree[b]++;
                }
            }
            int[] topolo = new int[n];
            int[] state = new int[n];
            for (int i = 0; i < n; i++) {
                if (degree[i] == 0 && state[i] == 0) {
                    boolean r = topo(state, i, topolo);
                    if (!r){
                        pw.println(-1);
                        return;
                    }
                }
            }
            if (idx != n) {
                pw.println(-1);
                return;
            }
            for (int i = 0; i < n; i++) {
                int u = topolo[i];
                if (nodes[u].edge.size() == 0) {
                    nodes[u].val[nodes[u].v]++;
                    continue;
                }
                for (int v: nodes[u].edge) {
                    for (int j = 0; j < 26; j++) {
                        nodes[u].val[j] = Math.max(nodes[u].val[j], nodes[v].val[j]);
                    }
                }
                nodes[u].val[nodes[u].v]++;
            }
            int t = 0;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < 26; j++) {
                    t = Math.max(t, nodes[i].val[j]);
                }
            }
            pw.println(t);
        }
        int idx = 0;
        public boolean topo(int[] state, int u, int[] stack){
            state[u] = 1;
            for (int v : nodes[u].edge) {
                if (state[v] == 1) {
                    return false;
                }
                if (state[v] == 0) {
                    boolean ret = topo(state, v, stack);
                    if (!ret) return false;
                }
            }
            state[u] = 2;
            stack[idx++] = u;
            return true;
        }



    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s){  br = new BufferedReader(new InputStreamReader(s));}

        public Scanner(FileReader s) throws FileNotFoundException {br = new BufferedReader(s);}

        public String next() throws IOException
        {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public int nextInt() throws IOException {return Integer.parseInt(next());}

        public long nextLong() throws IOException {return Long.parseLong(next());}

        public String nextLine() throws IOException {return br.readLine();}

        public double nextDouble() throws IOException { return Double.parseDouble(next()); }

        public boolean ready() throws IOException {return br.ready();}
    }
}
