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 Substring {
    static LinkedList<Integer>[] graph;
    static int n, m;
    static int[] reversePostOrder;
    static int cnt;
    static boolean[] visited;
    
    private static void dfs(int v) {
        visited[v] = true;
        for (int w: graph[v]) {
            if (!visited[w]) dfs(w);
        }
        reversePostOrder[cnt] = v;
        cnt--;
    }
    
    public static void main(String[] args) {
        //'a' is 97
        
        // I/O, construct graph
        Scanner in = new Scanner(System.in);
        n = in.nextInt();
        m = in.nextInt();
        in.nextLine();
        char[] letters = in.nextLine().toCharArray();
        graph = new LinkedList[n];
        for (int i = 0; i < n; i++) graph[i] = new LinkedList<>();
        
        for (int i = 0; i < m; i++) {
            int x = in.nextInt();
            int y = in.nextInt();
            graph[x-1].addFirst(y-1);
        }
        
        // top sort
        
        reversePostOrder = new int[n];
        cnt = n-1;
        visited = new boolean[n];
        for (int i = 0; i < n; i++) {
            if (!visited[i]) dfs(i);
        }
        
        int[][] dp = new int[n][26];
        visited = new boolean[n];
        int test;
        int curmax = 0;
        
        // dp, relaxing all edges from v when going forward at each stage
        
        for (int i = 0; i < n; i++) {
            int v = reversePostOrder[i];
            visited[v] = true; // this needs to be here to prevent self-loops
            // this is like relaxing the edge from v to the "finish line"
            test = dp[v][(int) letters[v] - 97] + 1;
            if (test > curmax) curmax = test;
            for (int w: graph[v]) {
                if (visited[w]) { // do we have a cycle?
                    System.out.println(-1);
                    System.exit(0);
                }
                else {
                    for (int c = 0; c < 26; c++) {
                        test = dp[v][c];
                        if (c == (int) letters[v] - 97) test++;
                        if (test > dp[w][c]) dp[w][c] = test;
                        if (test > curmax) curmax = test;
                    }
                }
            }
        }
        
        System.out.println(curmax);
    }
}