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

public class Abc {
    static int n, m;
    static ArrayList<Integer> adj[];
    static boolean vis[];
    static boolean inStack[];
    static char s[];
    static int dp[][];

    public static void main(String[] args) throws Exception {
        FastReader sc = new FastReader();
        n = sc.nextInt();
        m = sc.nextInt();
        s = sc.next().toCharArray();
        adj = new ArrayList[n];
        for (int i = 0; i < n; i++) adj[i] = new ArrayList<>();
        int indeg[] = new int[n];
        for (int i = 0; i < m; i++) {
            int a = sc.nextInt() - 1, b = sc.nextInt() - 1;
            adj[a].add(b);
            indeg[b]++;
        }
        vis = new boolean[n];
        inStack = new boolean[n];
        dp = new int[n][26];
        for (int i = 0; i < n; i++) {
            if (vis[i]) continue;
            if (cy(i)) {
                System.out.println(-1);
                return;
            }
        }
        Arrays.fill(vis, false);
        int max = 0;
        for (int i = 0; i < n; i++) {
            if (vis[i])continue;
            dfs(i);
        }
        for (int i=0;i<n;i++){
            for (int j=0;j<26;j++){
                max=Math.max(max,dp[i][j]);
            }
        }
        System.out.println(max);
    }

    static void dfs(int v) {
        vis[v]=true;
        for (int u:adj[v]){
            if (!vis[u]){
                dfs(u);
                for (int i=0;i<26;i++){
                    dp[v][i]=Math.max(dp[u][i],dp[v][i]);
                }
            }else {
                for (int i=0;i<26;i++){
                    dp[v][i]=Math.max(dp[u][i],dp[v][i]);
                }
            }
        }
        dp[v][s[v]-'a']++;
    }

    static boolean cy(int v) {
        inStack[v] = true;
        vis[v] = true;
        for (int u : adj[v]) {
            if (inStack[u]) return true;
            if (vis[u]) continue;
            if (cy(u)) return true;
        }
        inStack[v] = false;
        return false;
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        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());
        }

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

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}