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.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class DSubstring 
{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		
		//System.out.println("Enter n");
		int n=sc.nextInt();
		//System.out.println("Enter m");
		int m=sc.nextInt();
		
		//System.out.println("Enter s");
		String s=sc.next();
		
		ArrayList<Integer>x=new ArrayList<Integer>();
		
		ArrayList<Integer>y=new ArrayList<Integer>();
		
		for(int i=0;i<m;i++)
		{
		//	System.out.println("Enter x");
			
			int X=sc.nextInt();
			
			x.add(X);
			
		//	System.out.println("Enter y");
			
			int Y=sc.nextInt();
			
			y.add(Y);
		}
		
		
		sc.close();
		ArrayList<Integer> v[]=new ArrayList[n];
		
		for(int i=0;i<n;i++)
		{
			v[i]=new ArrayList<Integer>();
		}
		
		ArrayList<Integer>d=new ArrayList<Integer>();
		
		for(int i=0;i<n;i++)
		{
			d.add(0);
		}
		
		ArrayList<ArrayList<Long>>dp=new ArrayList<ArrayList<Long>>();
		for(int i=0;i<n;i++)
		{
			ArrayList<Long>temp=new ArrayList<Long>();
			for(int k=0;k<26;k++)
			{
				temp.add(0l);
			}
			dp.add(temp);
		}
		
		for(int i=0;i<m;i++)
		{
			x.set(i, x.get(i)-1);
			y.set(i,y.get(i)-1);
			
			v[x.get(i)].add(y.get(i));
			
			d.set(y.get(i),d.get(y.get(i))+1);
		}
		
		
		
		Queue<Integer>q=new LinkedList<Integer>();
		
		for(int i=0;i<n;i++)
		{
			if(d.get(i)==0l)
			{
				q.add(i);
			}
		}
		while(!q.isEmpty())
		{
			int k=q.remove();
			long temp=dp.get(k).get(s.charAt(k)-'a');
			
			dp.get(k).set(s.charAt(k)-'a',temp+1 );
			
			for(int i=0;i<v[k].size();i++)
			{
				int z=v[k].get(i);
				d.set(z, d.get(z)-1);
				for(int j=0;j<26;j++)
				{
					dp.get(z).set(j, Math.max(dp.get(z).get(j), dp.get(k).get(j)));
					
				}
				if(d.get(z)==0)
				{
					q.add(z);
				}
			}
			
		}
		Long ans=Long.MIN_VALUE;
		for(int i=0;i<n;i++)
		{
			if(d.get(i)!=0)
			{
				ans=Long.MIN_VALUE;
				break;
			}
			for(int j=0;j<26;j++)
			{
				ans=Math.max(ans, dp.get(i).get(j));
			}
		}
		if(ans!=Long.MIN_VALUE)
		{
			System.out.println(ans);
		}
		else
		{
			System.out.println(-1);
		}
	}

}
