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.*;
import java.util.ArrayList;
import java.util.StringTokenizer;


public class Main {
    static final int UNVIS = 0, EXPL = 1, VIS = 2;
    static ArrayList<Integer> adjList[];
    static int[] color, cnt[];
    static boolean cycle;
    static char[] c;
    static int ans;

    public static void main(String[] args) throws IOException {
        Scanner sc = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n = sc.nextInt(), m = sc.nextInt();
        c = sc.next().toCharArray();
        adjList = new ArrayList[n];
        for (int i = 0; i < n; i++)
            adjList[i] = new ArrayList<>();
        while (m-- > 0)
            adjList[sc.nextInt() - 1].add(sc.nextInt() - 1);
        color = new int[n];
        cnt = new int[n][26];
        for (int i = 0; i < n; i++)
            if (color[i] == UNVIS) dfs(i);
        out.println(cycle?-1:ans);
        out.flush();
        out.close();
    }

    static void dfs(int u) {
        if (cycle) return;
        color[u] = EXPL;
        for (int v : adjList[u]) {
            if (color[v] == EXPL)
                cycle = true;
            if (color[v] == UNVIS) dfs(v);
            for (int i = 0; i < 26; i++)
                cnt[u][i] = Math.max(cnt[v][i],cnt[u][i]);
        }
        cnt[u][c[u] - 'a']++;
        for (int x : cnt[u])
            ans = Math.max(ans, x);
        color[u] = VIS;
    }


    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream system) {
            br = new BufferedReader(new InputStreamReader(system));
        }

        public Scanner(String s) throws FileNotFoundException {

            br = new BufferedReader(new FileReader(new File(s)));
        }


        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

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

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

        public char nextChar() throws IOException {
            return next().charAt(0);
        }

        public Long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public boolean ready() throws IOException {
            return br.ready();
        }


        public int[] nextIntArray(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public long[] nextLongArray(int n) throws IOException {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }


        public Integer[] nextIntegerArray(int n) throws IOException {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public double[] nextDoubleArray(int n) throws IOException {
            double[] ans = new double[n];
            for (int i = 0; i < n; i++)
                ans[i] = nextDouble();
            return ans;
        }

        public short nextShort() throws IOException {
            return Short.parseShort(next());
        }

    }

}