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 accolite {

	static ArrayList<Integer>[] graph;
	static int[] letter;
	static int[] visted;
	static int[][] mem;
	private static int  beauty(int v, int e, String s, ArrayList<Integer> x, ArrayList<Integer> y) {
		// TODO Auto-generated method stub
		
		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=x.get(i);
			int to=y.get(i);
			graph[from].add(to);
		}
		
		letter =new int[v+1];
		
		for(int i=1;i<=v;i++)
		{
			letter[i]=s.charAt(i-1)-'a';
		}
		
		visted=new int[v+1];
		mem =new int[v+1][26];
		
		//
		for(int i=1;i<=v;i++)
		{
			if(visted[i]==0 && dfs(i))
			{
				return -1;
			}
		}
		int ans=0;
		for(int i=0;i<=v;i++)
			for(int j=0;j<26;j++)
				ans=Math.max(ans, mem[i][j]);
		
		
		return ans;
	}
	private static boolean dfs(int vertex) {
		visted[vertex]=1;
		for(int to:graph[vertex])
		{
			if(visted[to]==1)
				return true;
			if(visted[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]]++;
		visted[vertex]=2;
		return false;
	}
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		 Scanner sc=new Scanner(System.in);
		 
		 int n=sc.nextInt();
		 int m=sc.nextInt();
		 
		 String s=sc.next();
		 
		 ArrayList<Integer> x=new ArrayList<Integer>();	
		 ArrayList<Integer> y=new ArrayList<Integer>();
		 for(int i=0;i<m;i++)
		 {
			 x.add(sc.nextInt());
			 y.add(sc.nextInt());
		 }
//		 ArrayList<Integer> y=new ArrayList<Integer>();
//		 for(int i=0;i<m;i++)
//		 {
//			
//		 }
		 System.out.print(beauty(n,m,s,x,y));
	}

	

}
