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 Main {
    public static void main(String[] args) throws IOException {
        FastScanner in = new FastScanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        new Main().run(in, out);
        out.close();
    }


    List<Integer>[] adj;
    boolean[] visited;
    int[][] counts;
    char[] ca;

    int[] preorder;
    int preorderNum = 0;

    void run(FastScanner in, PrintWriter out) {

        int n = in.nextInt();
        int m = in.nextInt();
        int[] inDeg = new int[n];

        preorder = new int[n];

        adj = new List[n];
        visited = new boolean[n];
        counts = new int[n][26];

        for (int i = 0; i < n; i++) adj[i] = new ArrayList<>();

        ca = in.next().toCharArray();

        for (int i = 0; i < m; i++) {
            int u = in.nextInt()-1;
            int v = in.nextInt()-1;
            if (u == v) {
                out.println(-1);
                return;
            }
            adj[u].add(v);
            inDeg[v]++;
        }

        // get all inDeg 0 sources
        List<Integer> sources = new ArrayList<>();
        for (int i = 0; i < n; i++) if (inDeg[i] == 0) sources.add(i);
        // oops, all cycles
        if (sources.size() == 0) {
            out.println(-1);
            return;
        }

        int max = 0;
        for (int source : sources) {
            if (!go(source)) {
                out.println(-1);
                return;
            }
            for (int i = 0; i < 26; i++) {
                max = Math.max(max, counts[source][i]);
            }
        }

        // for (int i = 0; i < n; i++) {
        //     System.out.println(i + " " + Arrays.toString(counts[i]));
        // }

        out.println(max);
    }

    boolean go(int u) {

        if (visited[u]) return true;
        visited[u] = true;
        preorder[u] = preorderNum++;

        for (Integer v : adj[u]) {
            if (!go(v)) return false;
            for (int i = 0; i < 26; i++) {
                counts[u][i] = Math.max(counts[u][i], counts[v][i]);
            }

            if (preorder[v] <= preorder[u]) return false;

        }
        counts[u][ca[u]-'a']++;
        preorder[u] = Integer.MAX_VALUE;
        return true;
    }

    static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in));
            st = null;
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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