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.*;
import java.math.*;

public class Main4 {
	public static void main(String[] args) {
		try {
			System.setIn(new FileInputStream("input.txt"));
			System.setOut(new PrintStream(new FileOutputStream("output.txt")));
		} catch (Exception e) {
			System.err.println("Error");
		}
		Scanner scn = new Scanner(System.in);
		int v = scn.nextInt();
		int e = scn.nextInt();
		int[][] dp = new int[v][26];
		String str = scn.next();
		int[] arr = new int[v];

		for (int i = 0; i < v; i++) {
			arr[i] = str.charAt(i) - 'a';
		}
		List<Integer>[] graph = new ArrayList[v];

		for (int i = 0; i < v; i++) {
			graph[i] = new ArrayList<>();
		}
		for (int i = 0; i < e; i++) {
			int temp1 = scn.nextInt() - 1;
			int temp2 = scn.nextInt() - 1;
			graph[temp1].add(temp2);
		}

		int[] visited = new int[v];
		for (int i = 0; i < v; i++) {
			if (visited[i] == 0 && !beautiful(graph, dp, arr, i, visited)) {
				System.out.println("-1");
				return;
			}
		}

		int result = 0;
		for (int i = 0; i < v; i++) {
			for (int j = 0; j < 26; j++) {
				result = Math.max(dp[i][j], result);
			}
		}
		System.out.println(result);
	}

	public static boolean beautiful(List<Integer>[] graph, int[][] dp, int[] arr, int num, int[] visited) {
		visited[num] = 1;
		for (int temp : graph[num]) {
			if (visited[temp] == 1) return false;
			if (visited[temp] == 0 && !beautiful(graph, dp, arr, temp, visited)) {
				return false;
			}
			for (int i = 0; i < 26; i++) {
				dp[num][i] = Math.max(dp[num][i], dp[temp][i]);
			}
		}
		visited[num] = 2;
		dp[num][arr[num]]++;
		return true;
	}
}
