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 Substring {

	/************************ SOLUTION STARTS HERE ************************/

	static int adj[][];
	private static boolean visitedGlobal[];
	private static boolean visitedTemp[];

	private static boolean dfs(int u) {

		visitedGlobal[u] = true;
		visitedTemp[u] = true;
		for (int v : adj[u]) {

			if (visitedTemp[v])
				return true;

			if (!visitedGlobal[v]) {
				if (dfs(v))
					return true;
			}

		}
		visitedTemp[u] = false;
		return false;

	}

	private static boolean isCyclic(int V) {

		for (int i = 1; i <= V; i++) {

			if (!visitedGlobal[i]) {
				if (dfs(i))
					return true;
			}

		}

		return false;

	}

	static int memo[];
	static char str[];
	static char c;

	static int topDown(int u, char alph) {
		memo[u] = 0;
		for (int v : adj[u])
			memo[u] = Math.max(memo[u], memo[v] == -1 ? topDown(v, alph) : memo[v]);

		memo[u] += str[u - 1] == alph ? 1 : 0;
		return memo[u];
	}

	// Courtesy : UWI ( adjacency list using Jagged Arrays )
	static int[][] packD(int n, int[] from, int[] to, int isOneBased) {
		int[][] g = new int[n + isOneBased][];
		int[] p = new int[n + isOneBased];
		for (int f : from)
			p[f]++;
		for (int i = 0 + isOneBased; i < n + isOneBased; i++)
			g[i] = new int[p[i]];
		for (int i = 0; i < from.length; i++)
			g[from[i]][--p[from[i]]] = to[i];

		return g;
	}

	private static void solve() {

		int V = nextInt();
		int E = nextInt();

		str = nextLine().toCharArray();

		visitedGlobal = new boolean[V + 1];
		visitedTemp = new boolean[V + 1];

		int from[] = new int[E];
		int to[] = new int[E];

		while (E-- > 0) {
			int u = nextInt();
			int v = nextInt();
			from[E] = u;
			to[E] = v;
		}

		adj = packD(V, from, to, 1);

		if (isCyclic(V)) {
			println(-1);
			return;
		}

	     // Now a DAG
        memo = new int[V + 1];
        int max = 0;
        for(char ch = 'a'; ch <= 'z'; ch++) {
            Arrays.fill(memo, -1);
            for(int i = 1; i <= V; i++) {
                if(memo[i] == -1)
                    topDown(i, ch);
                max = Math.max(max , memo[i]);
            }
        }
        
        println(max);

	}

	/************************ SOLUTION ENDS HERE ************************/

	/************************ TEMPLATE STARTS HERE **********************/

	/*
	 * Increase stack size in java
	 * 
	 */
	public static void main(String[] args) throws IOException {
		new Thread(null, new Runnable() {
			public void run() {
				new Substring().run();
			}
		}, "Increase Stack", 1 << 25).start();

	}

	void run() {
		reader = new BufferedReader(new InputStreamReader(System.in));
		writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), false);
		st = null;
		solve();
		try {
			reader.close();
		} catch (IOException e) {
			e.printStackTrace();
		}
		writer.close();
	}

	static BufferedReader reader;
	static PrintWriter writer;
	static StringTokenizer st;

	static String next() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				String line = reader.readLine();
				if (line == null) {
					return null;
				}
				st = new StringTokenizer(line);
			} catch (Exception e) {
				throw new RuntimeException();
			}
		}
		return st.nextToken();
	}

	static String nextLine() {
		String s = null;
		try {
			s = reader.readLine();
		} catch (IOException e) {
			e.printStackTrace();
		}
		return s;
	}

	static int nextInt() {
		return Integer.parseInt(next());
	}

	static long nextLong() {
		return Long.parseLong(next());
	}

	static double nextDouble() {
		return Double.parseDouble(next());
	}

	static char nextChar() {
		return next().charAt(0);
	}

	static int[] nextIntArray(int n) {
		int[] a = new int[n];
		int i = 0;
		while (i < n) {
			a[i++] = nextInt();
		}
		return a;
	}

	static long[] nextLongArray(int n) {
		long[] a = new long[n];
		int i = 0;
		while (i < n) {
			a[i++] = nextLong();
		}
		return a;
	}

	static int[] nextIntArrayOneBased(int n) {
		int[] a = new int[n + 1];
		int i = 1;
		while (i <= n) {
			a[i++] = nextInt();
		}
		return a;
	}

	static long[] nextLongArrayOneBased(int n) {
		long[] a = new long[n + 1];
		int i = 1;
		while (i <= n) {
			a[i++] = nextLong();
		}
		return a;
	}

	static void print(Object o) {
		writer.print(o);
	}

	static void println(Object o) {
		writer.println(o);
	}

	/************************ TEMPLATE ENDS HERE ************************/

}