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.io.*;
import java.util.*;

public class D {

	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		out = new PrintWriter(new BufferedOutputStream(System.out));

		solve(sc, out);

		out.close();
	}

	private static void solve(FastScanner sc, PrintWriter out) {
		int n = sc.nextInt();
		int m = sc.nextInt();

		char nodes[] = sc.nextToken().toCharArray();

		int in[] = new int[n + 1];
		int count[][] = new int[n + 1][26];
		Map<Integer, List<Integer>> edge = new HashMap<Integer, List<Integer>>();

		for (int i = 0; i < m; i++) {
			int x = sc.nextInt();
			int y = sc.nextInt();
			in[y]++;
			if (!edge.containsKey(x)) {
				edge.put(x, new ArrayList<Integer>());
			}
			edge.get(x).add(y);
		}

		int ans = 0;
		int deep[] = new int[n + 1];
		int total = n;

		List<Integer> bfs = new LinkedList<Integer>();
		for (int i = 1; i <= n; i++) {
			if (in[i] == 0) {
				bfs.add(i);
				int c = nodes[i - 1] - 'a';
				count[i][c] = 1;
				deep[i] = 1;
				ans = 1;
			}
		}

		while (!bfs.isEmpty()) {
			int current = bfs.remove(0);
			total--;
			if (!edge.containsKey(current)) {
				continue;
			}
			List<Integer> edges = edge.get(current);
			Iterator<Integer> it = edges.iterator();
			while (it.hasNext()) {
				int next = it.next();
				int nextc = nodes[next - 1] - 'a';
				for (int i = 0; i < 26; i++) {
					int newValue = count[current][i] + (i == nextc ? 1 : 0);
					if (newValue > count[next][i]) {
						count[next][i] = newValue;
						ans = Math.max(ans, newValue);
					}
				}
				in[next]--;
				if (in[next] == 0) {
					bfs.add(next);
				}
			}
		}
		if (total != 0) {
			ans = -1;
		}
		out.println(ans);
	}

	// -----------PrintWriter for faster output---------------------------------
	public static PrintWriter out;

	// -----------MyScanner class for faster input----------
	public static class FastScanner {
		BufferedReader br;
		StringTokenizer st;

		public FastScanner(String s) {
			try {
				br = new BufferedReader(new FileReader(s));
			} catch (FileNotFoundException e) {
				// TODO Auto-generated catch block
				e.printStackTrace();
			}
		}

		public FastScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}

		String nextToken() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					// TODO Auto-generated catch block
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		int nextInt() {
			return Integer.parseInt(nextToken());
		}

		long nextLong() {
			return Long.parseLong(nextToken());
		}

		double nextDouble() {
			return Double.parseDouble(nextToken());
		}
	}
	// --------------------------------------------------------
}