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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.util.ArrayDeque;
import java.util.Collections;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author Atharva Nagarkar
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader in = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        DSubstring solver = new DSubstring();
        solver.solve(1, in, out);
        out.close();
    }

    static class DSubstring {
        int n;
        int m;
        char[] s;
        int[] a;
        ArrayList<Integer>[] vs;
        ArrayList<Integer> order;
        int[] indeg;

        boolean cycle() {
            int cnt = 0;
            ArrayDeque<Integer> q = new ArrayDeque<>();
            for (int i = 0; i < n; ++i) {
                if (indeg[i] == 0) {
                    q.addLast(i);
                }
            }
            while (!q.isEmpty()) {
                int cur = q.pollFirst();
                order.add(cur);
                ++cnt;
                for (int i : vs[cur]) {
                    indeg[i]--;
                    if (indeg[i] == 0) {
                        q.addLast(i);
                    }
                }
            }
            return cnt != n;
        }

        public void solve(int testNumber, InputReader in, PrintWriter out) {
            n = in.nextInt();
            m = in.nextInt();
            s = in.next().toCharArray();
            a = new int[n];
            vs = new ArrayList[n];
            indeg = new int[n];
            order = new ArrayList<>();
            for (int i = 0; i < n; ++i) {
                vs[i] = new ArrayList<>();
                a[i] = s[i] - 'a';
            }
            for (int i = 0; i < m; ++i) {
                int u = in.nextInt() - 1;
                int v = in.nextInt() - 1;
                vs[u].add(v);
                ++indeg[v];
            }
            if (cycle()) {
                out.println(-1);
                return;
            }
            Collections.reverse(order);
            int[][] dp = new int[26][n];
            int ans = 1;
            for (int node : order) {
                dp[a[node]][node]++;
                int[] cur = new int[26];
                for (int adj : vs[node]) {
                    for (int c = 0; c < 26; ++c) {
                        cur[c] = Math.max(cur[c], dp[c][adj]);
                    }
                }
                for (int c = 0; c < 26; ++c) {
                    dp[c][node] += cur[c];
                    ans = Math.max(ans, dp[c][node]);
                }
            }
            out.println(ans);
        }

    }

    static class InputReader {
        public BufferedReader reader;
        public StringTokenizer tokenizer;

        public InputReader(InputStream stream) {
            reader = new BufferedReader(new InputStreamReader(stream), 32768);
            tokenizer = null;
        }

        public String next() {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (IOException e) {
                    throw new RuntimeException(e);
                }
            }
            return tokenizer.nextToken();
        }

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

    }
}

