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.*;

public class Main {
	static int MAX_E = 300000;
	static int MAX_ALPH = 26;
	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(); int m = sc.nextInt();
		String g = sc.next();
		ArrayList<ArrayList<Integer>> es = new ArrayList<>(m);
		int[] deg = new int[n];
		for(int i=0; i<n; i++) {
			es.add(new ArrayList<Integer>());
		}
		for(int i=0; i<m; i++) {
			int from = sc.nextInt()-1; int to = sc.nextInt()-1;
			es.get(from).add(to);
			deg[to]++;
		}
		int[][] f = new int[n][MAX_ALPH];
		Queue<Integer> q = new ArrayDeque<Integer>();
		for(int i=0; i<n; i++) {
			if(deg[i]==0) {
				q.add(i);
				f[i][ord(g.charAt(i))]++;
			}
		}
		int cnt = 0;
		while(!q.isEmpty()) {
			int from = q.poll();
			for(int to : es.get(from)) {
				for(int i=0; i<MAX_ALPH; i++) {
					if(ord(g.charAt(to))==i) {
						f[to][i] = Math.max(f[to][i], f[from][i]+1);
					} else {
						f[to][i] = Math.max(f[to][i], f[from][i]);
					}
				}
				deg[to]--;
				if(deg[to]==0) {
					q.add(to);
				}
			}
			cnt++;
		}
		if(cnt!=n) {
			System.out.println(-1); return;
		}
		int ans = 0;
		for(int i=0; i<n; i++) {
			for(int j=0; j<MAX_ALPH; j++) ans = Math.max(ans, f[i][j]);
		}
		System.out.println(ans);
	}
	
	static int ord(char ch) {
		return ch - 'a';
	}
	
	static int binarySearch(int[] A, int x) {
		int l = 0;
		int r = A.length;
		while(l<r) {
			int mid = (l+r)/2;
			if(A[mid] == x) return mid;
			else if(x<A[mid]) r = mid;
			else l = mid+1;
		}
		return -1;
	}
	
	static long upper_bound(int[] A, int b) {
		// bよりおおきい値がはじめて出てくるindexを返す
		return lower_bound(A, b+1);
	}

	static long lower_bound(int[] A, int b) {
		// b以上の値がはじめて出てくるindexを返す
		int lb=0;
		int ub=A.length;
		while(ub>lb) {
			int mid = (lb+ub)/2;
			if(A[mid]>=b) ub = mid;
			else lb = mid+1;
		}
		return lb;
	}
	
	// union find lib
	// usage:
	// 最初にinitを呼ぶ
	// root: 直接は呼ばないで
	// unite: まとめる
	// same: グループ判定
	static void init(int par[], int N) {
		for(int i=0; i<N; i++) {
			par[i] = i;
		}
	}
	
	static int root(int x, int [] par) {
		if(par[x] == x) {
			return x;
		} else {
			return (par[x] = root(par[x], par));
		}
	}
	
	static boolean same(int x, int y, int[] par) {
		return root(x, par) == root(y, par);
	}
	
	static void unite(int x, int y, int[] par) {
		x = root(x, par);
		y = root(y, par);
		if(x == y) return;
		par[x] = y;
	}
	
	// end union find lib
	
	// number lib
	
	static int max(int...ls) {
		int ans = Integer.MIN_VALUE;
		for(int i=0; i<ls.length; i++) {
			ans = Math.max(ans, ls[i]);
		}
		return ans;
	}
	
	static int min(int...ls) {
		int ans = Integer.MAX_VALUE;
		for(int i=0; i<ls.length; i++) {
			ans = Math.min(ans, ls[i]);
		}
		return ans;
	}
	
	static long lcm(long a, long b) {
		return a*(b/gcd(a, b));
	}
	
	static long gcd(long a, long b) {
		long ta = Math.max(a, b);
		long tb = Math.min(a, b);
		long tmp;
		while((tmp = ta%tb) != 0) {
			ta = tb;
			tb = tmp;
		}
		return tb;
	}
	
	static long modcomb(long n, long k, long mod) {
		if(k==1) {
			return n;
		}
		
		long ans = 1;
		for(long i=n; i>=n-k+1; i--) {
			ans = (ans * i)%mod;
		}
		for(long i=k; 0<i; i--) {
			ans = (ans * modpow(i, mod-2, mod)) % mod;
		}
		return ans;
	}
	
	static long modpow(long a, long b, long mod) {
		if(b==0) return 1;
		if(b%2==0) {
			long d = modpow(a, b/2, mod);
			return (d*d)%mod;
		} else {
			return (a*modpow(a, b-1, mod))%mod;
		}
	}
	
	static int disit(long a, long d) {
		int count = 0;
		while(a!=0) {
			a = a/d;
			count++;
		}
		return count;
	}
	
	// end number lib
}