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.*;

public class acco {
    static int n, m, freq[], res;
    static String st;
    static List<Integer> adj[];
    static int[][] dp;
    static boolean[] visited;
    static Set<Integer> set = new HashSet<Integer>();
    public static void dfs(int u) {
        if(res == -1){
            return;
        }
        visited[u] = true;
        dp[u][st.charAt(u) - 'a']++;
        set.add(u);
        for (int i = 0; i < adj[u].size(); i++) {
            int x = adj[u].get(i);
            if(set.contains(x)){
                res = -1;
                return;
            }
            if (!visited[x]) {
                dfs(x);
            }
        }
        for (int j = 0; j <= 'z' - 'a'; j++) {
            int max = 0;
            for (int i = 0; i < adj[u].size(); i++) {
                int x = adj[u].get(i);
                if(set.contains(x)){
                    res = -1;
                    return;
                }
                max = Math.max(dp[x][j], max);
            }
            dp[u][j] += max;
        }
        set.remove(u);
    }
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        String s = sc.next();
        int x[] = new int[m];
        int y[] = new int[m];
        for (int i = 0; i < m; i++) {
            x[i] = sc.nextInt();
            y[i] = sc.nextInt();
        }
        System.out.println(beauty(n,m,s,x,y));
    }
    public static  int beauty(int n1, int m1,String str, int x[], int y[]){
        n= n1;
        m=m1;
        st=str;
        adj = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            adj[i] = new ArrayList<Integer>();
        }
        freq = new int[27];
        dp = new int[n][27];
        visited = new boolean[n];
        for (int i = 0; i < m; i++) {
            int u = x[i] - 1;
            int v = y[i]- 1;
            adj[u].add(v);
        }
        res = 0;
        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                dfs(i);
            }
        }
        if(res == -1){
            return -1;
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j <= 'z' - 'a'; j++) {
                res = Math.max(dp[i][j], res);
            }
        }
        return res;
    }

}