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 D {

	private static ArrayList<Integer>[] graph;
	private static int[] visited;
	private static int[] letter;
	private static int[][] dp;
	private static int V, E;

	public static void main(String... args) throws Exception {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer(br.readLine());
		V = Integer.parseInt(st.nextToken());
		E = Integer.parseInt(st.nextToken());

		String path = br.readLine();
		letter = new int[V + 1];
		for (int i = 1; i <= V; i++)
			letter[i] = path.charAt(i - 1) - 'a';

		graph = new ArrayList[V + 1];
		for (int i = 0; i <= V; i++)
			graph[i] = new ArrayList<Integer>();

		for (int i = 0; i < E; i++) {
			st = new StringTokenizer(br.readLine());
			graph[Integer.parseInt(st.nextToken())].add(Integer.parseInt(st.nextToken()));
		}

		dp = new int[V + 1][26];
		visited = new int[V + 1];
		for (int i = 1; i <= V; i++) {
			if (visited[i] == 0 && dfs(i)) {
				System.out.print(-1);
				System.exit(0);
			}
		}

		int answer = 0;
		for (int i = 1; i <= V; i++)
			for (int j = 0; j < 26; j++)
				answer = Math.max(answer, dp[i][j]);
		System.out.print(answer);
	}

	private static boolean dfs(int vertex) {
		visited[vertex] = 1;

		for (int to : graph[vertex]) {
			if (visited[to] == 1)
				return true;
			if (visited[to] == 0 && dfs(to))
				return true;

			for (int i = 0; i < 26; i++)
				dp[vertex][i] = Math.max(dp[vertex][i], dp[to][i]);
		}

		dp[vertex][letter[vertex]]++;
		visited[vertex] = 2;
		return false;
	}
}