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.*;
import java.io.*;
import java.math.*;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
public class Main {

  static int []deg = new int[300005];
  static int [][]dp = new int[300005][26];
  static ArrayList<Integer> a[] = new ArrayList[300005];
  static int []str;// = new int[26];


  public static void main(String[] args) {
    try {
      MScanner in = new MScanner(System.in);
      //BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out, "UTF-8"));
      int n = in.nextInt();
      int m = in.nextInt();
      str = new int[n];
      for (int i = 0; i <= n; i++) {
        a[i] = new ArrayList<>();
      }

      String s0 = in.next();
      for (int j = 0;j < s0.length();++j) {
        char c = s0.charAt(j);
        str[j] = (int)(c - 'a');
      }
      for (int i = 0;i < m;++i) {
        int x = in.nextInt() - 1;
        int y = in.nextInt() - 1;
        a[y].add(x);
        deg[x]++;
      }
      Queue<Integer> Q = new LinkedList<Integer>();
      for (int i = 0;i < n;++i) {
        if (deg[i] == 0) Q.add(i);
      }
      while (Q.size() > 0) {
        int p = Q.poll();
        dp[p][str[p]] += 1;
        for (int u:a[p]) {
          for (int i = 0;i < 26;++i) dp[u][i] = Math.max(dp[u][i], dp[p][i]);
          deg[u]--;
          if (deg[u] == 0) Q.add(u);
        }
      }
      int ans = 0;
      for (int i=0;i<n;i++) {
        if (deg[i] > 0) {
          System.out.println(-1);
          return ;
        }
        for (int j=0;j<26;j++) ans = Math.max(ans,dp[i][j]);
      }
      System.out.println(ans);

    } catch (Exception e) {
      e.printStackTrace();
    }
  }
  static class MScanner {
    StringTokenizer st;
    BufferedReader br;
    public MScanner(InputStream system) {
      br = new BufferedReader(new InputStreamReader(system));
    }

    public MScanner(String file) throws Exception {
      br = new BufferedReader(new FileReader(file));
    }

    public String next() throws IOException {
      while (st == null || !st.hasMoreTokens())
	st = new StringTokenizer(br.readLine());
      return st.nextToken();
    }
    public int[] takearr(int n) throws IOException {
      int[]in=new int[n];for(int i=0;i<n;i++)in[i]=nextInt();
      return in;
    }
    public long[] takearrl(int n) throws IOException {
      long[]in=new long[n];for(int i=0;i<n;i++)in[i]=nextLong();
      return in;
    }
    public Integer[] takearrobj(int n) throws IOException {
      Integer[]in=new Integer[n];for(int i=0;i<n;i++)in[i]=nextInt();
      return in;
    }
    public Long[] takearrlobj(int n) throws IOException {
      Long[]in=new Long[n];for(int i=0;i<n;i++)in[i]=nextLong();
      return in;
    }
    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 void waitForInput() throws InterruptedException {
      Thread.sleep(3000);
    }
  }
}


