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.
// practice with kaiboy
import java.io.*;
import java.util.*;

public class CF919D extends PrintWriter {
	CF919D() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF919D o = new CF919D(); o.main(); o.flush();
	}

	static final int A = 26;
	int[] kk = new int[A];
	int[] oo; int[][] oj;
	void link(int i, int j) {
		int o = oo[i];
		if (o >= 2 && (o & o - 1) == 0)
			oj[i] = Arrays.copyOf(oj[i], o << 1);
		oj[i][oo[i]++] = j;
	}
	int[] qu; int cnt = 0;
	byte[] used, cc;
	void init(int n) {
		oo = new int[n]; oj = new int[n][2];
		qu = new int[n];
		used = new byte[n];
	}
	boolean dfs(int i) {
		if (used[i] != 0)
			return used[i] == 1;
		used[i] = 1;
		for (int o = 0; o < oo[i]; o++) {
			int j = oj[i][o];
			if (dfs(j))
				return true;
		}
		used[i] = 2;
		qu[cnt++] = i;
		return false;
	}
	void main() {
		int n = sc.nextInt();
		int m = sc.nextInt();
		init(n);
		cc = sc.next().getBytes();
		while (m-- > 0) {
			int i = sc.nextInt() - 1;
			int j = sc.nextInt() - 1;
			link(i, j);
		}
		for (int i = 0; i < n; i++)
			if (used[i] == 0 && dfs(i)) {
				println(-1);
				return;
			}
		int[][] dp = new int[n][A];
		int ans = 0;
		for (int h = n - 1; h >= 0; h--) {
			int i = qu[h];
			int a = cc[i] - 'a';
			dp[i][a]++;
			for (a = 0; a < A; a++)
				ans = Math.max(ans, dp[i][a]);
			for (int o = 0; o < oo[i]; o++) {
				int j = oj[i][o];
				for (a = 0; a < A; a++)
					dp[j][a] = Math.max(dp[j][a], dp[i][a]);
			}
		}
		println(ans);
	}
}
