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.
//package codeforces.R460;


import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class D_
{
	static int n, m;
	static int[][] adj;
	static char[] nodes;
	public static void main(String[] args) throws IOException
	{
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		
		n = sc.nextInt();
		m = sc.nextInt();
		nodes =sc.next().toCharArray();
		vis = new int[n];
		int[] from = new int[m];
		int[] to = new int[m];
		for (int i = 0; i < m; i++)
		{
			int a = sc.nextInt()-1;
			int b = sc.nextInt()-1;
			from[i] = a; to[i] = b;
		}
		
		adj = packD(n, from, to);
		for (int i = 0; i < n; i++)
			if(vis[i] == 0)
				dfs(i);
		
		if(cycle)
		{
			System.out.println(-1);
			return;
		}
		
		vid = 2;
		ans = new int[n];
		for (int i = 0; i < 26; i++)
		{
			vid++;
			cur = (char)('a' + i);
			for (int j = 0; j < n; j++)
			{
				if(vis[j] != vid)
					solve(j);
			}
		}
		
		System.out.println(mx);
		
		pw.flush();
		pw.close();
	}
	static int vid;
	static boolean cycle = false;
	static int[] vis;
	static char cur = 'a';
	static int[] ans;
	static void dfs(int u)
	{
		vis[u] = 1;
		for (int v : adj[u])
		{
			if(vis[v] == 1)
				cycle = true;
			else if(vis[v] == 0)
				dfs(v);
		}
		vis[u] = 2;
	}
	
	static int mx = 0;
	static void solve(int u)
	{
		ans[u] = 0;
		vis[u] = vid;
		if(nodes[u] == cur)
			ans[u]++;
		int max = 0;
		for (int v : adj[u])
		{
			if(vis[v] != vid)
				solve(v);
			max = Math.max(max, ans[v]);
		}
		
		ans[u] += max;
		mx = Math.max(mx, ans[u]);
	}
	
	static int[][] packD(int n, int[] from, int[] to) {
		int[][] g = new int[n][];
		int[] p = new int[n];
		for (int f : from)
			p[f]++;
		for (int i = 0; i < n; i++)
			g[i] = new int[p[i]];
		for (int i = 0; i < from.length; i++) {
			g[from[i]][--p[from[i]]] = to[i];
		}
		return g;
	}
	
	static class Scanner 
	{
		StringTokenizer st; BufferedReader br;
		public Scanner(InputStream s){	br = new BufferedReader(new InputStreamReader(s));}
		public String next() throws IOException {while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());return st.nextToken();}
		public int nextInt() throws IOException {return Integer.parseInt(next());}
		public long nextLong() throws IOException {return Long.parseLong(next());}
		public String nextLine() throws IOException {return br.readLine();}
		public boolean ready() throws IOException {return br.ready();}
	}
}
