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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class D919 {
    private static int[][] maxCount;
    private static int[] mapping;

    private static List<List<Integer>> adjList;
    private static boolean[] visited;

    private static boolean[] vertexInPath;

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st;

        st = new StringTokenizer(br.readLine());
        int n = Integer.parseInt(st.nextToken());
        int m = Integer.parseInt(st.nextToken());

        mapping = new int[n];
        String x = br.readLine();
        for (int i = 0; i < n; i++) {
            mapping[i] = x.charAt(i) - 'a';
        }

        maxCount = new int[n][26];
        adjList = new ArrayList<>(n);
        for (int i = 0; i < n; i++) {
            adjList.add(new ArrayList<>());
        }
        for (int i = 0; i < m; i++) {
            st = new StringTokenizer(br.readLine());
            adjList.get(Integer.parseInt(st.nextToken()) - 1)
                    .add(Integer.parseInt(st.nextToken()) - 1);
        }

        visited = new boolean[n];
        vertexInPath = new boolean[n];
        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                count(i);
            }
        }
        int max = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < 26; j++) {
                max = Math.max(max, maxCount[i][j]);
            }
        }
        System.out.println(max);
    }

    private static void count(int u) {
        visited[u] = true;
        vertexInPath[u] = true;
        for (int v : adjList.get(u)) {
            if (vertexInPath[v]) {
                System.out.println(-1);
                System.exit(0);
            }
            if (!visited[v]) {
                count(v);
            }
            for (int i = 0; i < 26; i++) {
                maxCount[u][i] = Math.max(maxCount[u][i], maxCount[v][i]);
            }
        }
        maxCount[u][mapping[u]]++;
        vertexInPath[u] = false;
    }
}
