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.
input=lambda : stdin.readline().strip()
from sys import stdin, stdout 

def INI():
	return int(stdin.readline())
	
def INL():
	return [int(_) for _ in stdin.readline().split()]
	
def INS():
	return stdin.readline()

def MOD():
    return pow(10,9)+7
	
def OPS(ans):
	stdout.write(str(ans)+"\n")
	
def OPL(ans):
	[stdout.write(str(_)+" ") for _ in ans]
	stdout.write("\n")

	
from collections import deque
if __name__=="__main__":
	n,m=INL()
	s=input()
	x,y=[],[]
	for _ in range(m):
		a,b=map(int,input().split())
		x.append(a)
		y.append(b)
	
	l=list(s)
	D=[0 for _ in range(n)]
	G={_:[] for _ in range(n)}
	for _ in range(m):
		a,b=x[_],y[_]
		a-=1
		b-=1
		G[a].append(b)
		D[b]+=1
	q=deque()
	for _ in range(n):
		if D[_]==0:
			q.append(_)
	
	c=0
	ans=0
	DP=[[0 for _ in range(26)] for _ in range(n)]
	while c<n and q:
		x=q.popleft()
		c+=1
		DP[x][ord(l[x])-97]+=1
		for _ in G[x]:
			for __ in range(26):
				DP[_][__]=max(DP[_][__],DP[x][__])
			D[_]-=1
			if D[_]==0:
				q.append(_)
				
	if c!=n:
		OPS(-1)
	else:
		for _ in range(n):
			ans=max(ans,max(DP[_]))
		OPS(ans)