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.*;

public class File {
	
	public static class FastScanner {
		BufferedReader br;
		StringTokenizer st;
		
		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		
		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				}
				catch (IOException e) {
					e.printStackTrace();
				}
			}
			
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
		
		
		int n = sc.nextInt(); // Nodes
		int m = sc.nextInt(); // Directed edges
		
		String s = sc.next();
		
		Map<Integer, List<Integer>> graph = new HashMap<>();
		int[] indegree = new int[n];
		
		for (int i = 0; i < m; i++) {
			int u = sc.nextInt()-1;
			int v = sc.nextInt()-1;
			
			if (!graph.containsKey(u)) {
				graph.put(u, new ArrayList<>());
			}
			
			graph.get(u).add(v);
			indegree[v]++;
		}
		
		int processed = 0;
		LinkedList<Integer> queue = new LinkedList<>();
		
		for (int i = 0; i < n; i++) {
			if (indegree[i] == 0) {
				queue.add(i);
				processed++;
			}
		}
		
		int[][] dp = new int[n][26];
		while (!queue.isEmpty()) {
			int curr = queue.removeFirst();
			dp[curr][s.charAt(curr)-'a']++;
			
			for (int next : graph.getOrDefault(curr, new ArrayList<>())) {
				for (int c = 0; c < 26; c++) {
					dp[next][c] = Math.max(dp[next][c], dp[curr][c]);
				}
				
				indegree[next]--;
				
				if (indegree[next] == 0) {
					queue.addLast(next);
					processed++;
				}
			}
		}
		
		if (processed != n) { // There is a cycle.
			out.println(-1);
		}
		else {
			int max = 0;
			
			for (int i = 0; i < n; i++) {
				for (int j = 0; j < 26; j++) {
					max = Math.max(max, dp[i][j]);
				}
			}
			
			out.println(max);
		}
		
		out.close();
	}
}