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 ACC2
{

	static ArrayList<Integer> []graph;
	static int [] letter;
	static int [] visited;
	static int [][] mem;
   public static void main(String[] args) throws Exception {
   	Scanner obj = new Scanner(System.in);
   	int v = obj.nextInt();
   	int e = obj.nextInt();
   	String s = obj.next();
   	graph = new ArrayList[v+1];
   	for(int i=0;i<=v;i++)
   		 graph[i] = new ArrayList<Integer>();
   
   for(int i=0;i<e;i++)
   	{
   		int from = obj.nextInt();
   		int to = obj.nextInt();
   		graph[from].add(to);
   	}
   letter = new int[v+1];
   for(int i=1;i<=v;i++)
   	  letter[i] = s.charAt(i-1)-'a';
   visited = new int[v+1];
      mem = new int[v+1][26];

   for(int i=1;i<=v;i++)
   {
     if(visited[i]==0&&dfs(i))
     	 {
            System.out.println("-1");
            System.exit(0);
            
     	 }
   }
   int answer = 0;
   for(int i=0;i<=v;i++)
   	 for(int j=0;j<26;j++)
   	 	answer = Math.max(answer,mem[i][j]);

   	 System.out.println(answer);
}


public static boolean dfs(int vertex)
{

   visited[vertex] = 1;
   for(int to:graph[vertex])
   {
     if(visited[to]==1)
     	 return true;
     if(visited[to]==0&&dfs(to))
     	 return true;
    for(int i=0;i<26;i++)
    	mem[vertex][i] = Math.max(mem[vertex][i],mem[to][i]);

   }

   mem[vertex][letter[vertex]]++;
   visited[vertex]=2;
   return false;

}
}