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.lang.*;
import java.io.*;
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;

public class Main {
    char s[];
    ArrayList<Integer> g[];
    int used[];
    boolean clr[];
    void dfs11(int v, int type){
        used[v] = type;
        for(int u : g[v]){
            if(used[u] == type){
                if(!clr[u]) {
                    f = true;
                }
            }else if(used[u] == 0){
                dfs11(u, type);
            }
        }
        clr[v] = true;
    }
    void dfs(int v){
        used[v] = 1;
        for(int u : g[v]){
            if(used[u] == 0) dfs(u);
            for (int i = 0; i < 26; i++) {
                len[i][v] = Math.max(len[i][v], len[i][u]);
            }
        }
        len[s[v] - 'a'][v]++;
        for (int i = 0; i < 26; i++) {
            res = Math.max(res, len[i][v]);
        }
    }
    boolean f = false;
    boolean end[];
    int len[][];
    int res = 0;
    public void solve() throws IOException {
        int n = nextInt(), m = nextInt();
        s = nextToken().toCharArray();
        g = new ArrayList[n];
        used = new int[n];
        end = new boolean[n];
        len = new int[26][n];
        for (int i = 0; i < n; i++) {
            g[i] = new ArrayList<>();
        }
        for (int i = 0; i < m; i++) {
            int x = nextInt() - 1, y = nextInt() - 1;
            end[y] = true;
            g[x].add(y);
        }
        int type = 1;
        clr=  new boolean[n];
        for (int i = 0; i < n; i++) {
            if(!clr[i]){
                dfs11(i, type);
                type++;
            }
        }
        if(f){
            out.print(-1);
            return;
        }
        used = new int[n];
        for (int i = 0; i < n; i++) {
            if(used[i] == 0){
                dfs(i);
            }
        }
        out.print(res);
    }

    BufferedReader br;
    StringTokenizer sc;
    PrintWriter out;

    public static void main(String[] args) throws IOException {
        Locale.setDefault(Locale.US);
        new Main().run();
    }

    void run() throws IOException {
        try {
            br = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
//            br = new BufferedReader(new FileReader("pnumbers.in"();
//            out = new PrintWriter(new File("out.txt"));
            solve();
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(1);
        }
    }

    String nextToken() throws IOException {
        while (sc == null || !sc.hasMoreTokens()) {
            try {
                sc = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                return null;
            }
        }
        return sc.nextToken();
    }

    int nextInt() throws IOException {
        return Integer.parseInt(nextToken());
    }

    long nextLong() throws IOException {
        return Long.parseLong(nextToken());
    }

    double nextDouble() throws IOException {
        return Double.parseDouble(nextToken());
    }
}