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.util.StringTokenizer;
import java.io.BufferedReader;
import java.util.ArrayDeque;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
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);
        TaskD solver = new TaskD();
        solver.solve(1, in, out);
        out.close();
    }

    static class TaskD {
        public void solve(int testNumber, InputReader in, PrintWriter out) {
            int n = in.nextInt();
            int m = in.nextInt();
            String s = in.next();
            int[] inDeg = new int[n + 1];
            ArrayList<ArrayList<Integer>> dgr = new ArrayList<>();
            for (int i = 0; i <= n; i++) dgr.add(new ArrayList<>());
            for (int i = 0; i < m; i++) {
                int u = in.nextInt();
                int v = in.nextInt();
                dgr.get(u).add(v);
                inDeg[v]++;
            }
            ArrayDeque<Integer> noInEdge = new ArrayDeque<>();
            int[][] dp = new int[n + 1][26];
            for (int i = 1; i <= n; i++) {
                if (inDeg[i] == 0) noInEdge.add(i);
                dp[i][s.charAt(i - 1) - 'a'] = 1;
            }
            int rem = n;
            int maxV = 0;
            while (!noInEdge.isEmpty()) {
                rem--;
                int u = noInEdge.removeFirst();
                for (int v : dgr.get(u)) {
                    for (int i = 0; i < 26; i++) {
                        dp[v][i] = Math.max(dp[v][i], dp[u][i] + ((s.charAt(v - 1) == 'a' + i) ? 1 : 0));
                        maxV = Math.max(maxV, dp[v][i]);
                    }
                    inDeg[v]--;
                    if (inDeg[v] == 0) noInEdge.add(v);
                }
            }
            out.println((rem > 0) ? -1 : maxV);
        }

    }

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

        public InputReader(InputStream inputStream) {
            reader = new BufferedReader(new InputStreamReader(inputStream));
        }

        private void fillTokenizer() {
            if (tokenizer == null || !tokenizer.hasMoreTokens()) {
                try {
                    tokenizer = new StringTokenizer(reader.readLine());
                } catch (Exception e) {
                    throw new RuntimeException(e);
                }
            }
        }

        public String next() {
            fillTokenizer();
            return tokenizer.nextToken();
        }

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

    }
}

