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.*;
import java.util.*;

public class Substring {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        InputReader inp = new InputReader(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        Solver solver = new Solver();
        solver.solve(inp, out);
        out.close();
    }

    static class Solver {
        int[][] g, dp;
        int[] in_degree, out_degree;
        char[] s;
        boolean[] visited;

        private void dfs(int node) {
            if (visited[node]) return;
            visited[node] = true;
            for (int neigh: g[node]) {
                for (int i = 0; i < 26; i++) {
                    int newValue = dp[node][i];
                    if (s[neigh] - 97 == i) newValue++;
                    dp[neigh][i] = Math.max(dp[neigh][i], newValue);
                } if (--in_degree[neigh] == 0 && !visited[neigh]) dfs(neigh);
            }
        }

        private void solve(InputReader inp, PrintWriter out) {
            int n = inp.nextInt(), m = inp.nextInt();
            s = inp.next().toCharArray();
            g = new int[n][];
            dp = new int[n][26];
            in_degree = new int[n];
            out_degree = new int[n];
            visited = new boolean[n];
            int[][] edges = new int[m][2];
            for (int i = 0; i < m; i++) {
                int from = inp.nextInt() - 1, to = inp.nextInt() - 1;
                out_degree[from]++;
                in_degree[to]++;
                edges[i][0] = from;
                edges[i][1] = to;
            } for (int i = 0; i < n; i++) g[i] = new int[out_degree[i]];
            int[] cnt = new int[n];
            for (int i = 0; i < m; i++) {
                int[] edge = edges[i];
                g[edge[0]][cnt[edge[0]]] = edge[1];
                cnt[edge[0]]++;
            } for (int i = 0; i < n; i++) dp[i][s[i] - 97]++;
            for (int i = 0; i < n; i++) if (in_degree[i] == 0 && !visited[i]) dfs(i);
            boolean loop = false;
            int max = 0;
            for (int i = 0; i < n; i++) {
                for (int j = 0; j < 26; j++) max = Math.max(max, dp[i][j]);
                if (in_degree[i] > 0) loop = true;
            } out.print(loop ? -1: max);
        }
    }

    static class InputReader {
        BufferedReader reader;
        StringTokenizer tokenizer;

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

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