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.
from sys import stdin
input=lambda : stdin.readline().strip()
from math import ceil,sqrt,factorial,gcd
from collections import deque
from bisect import bisect_left
n,m=map(int,input().split())
s=input()
x=[]
y=[]
for i in range(m):
    a,b=map(int,input().split())
    x.append(a)
    y.append(b)

def beauty(n,m,s,x,y):
    l=list(s)
    degree=[0 for i in range(n)]
    graph={i:[] for i in range(n)}
    for i in range(m):
        a,b=x[i],y[i]
        a-=1
        b-=1
        graph[a].append(b)
        degree[b]+=1
    q=deque()
    for i in range(n):
        if degree[i]==0:
            q.append(i)
    
    count=0
    ans=0
    # print(degree)
    dp=[[0 for i in range(26)] for i in range(n)]
    while count<n and q:
            x=q.popleft()
            count+=1
            # print(ord(l[x])-97)
            dp[x][ord(l[x])-97]+=1
            for i in graph[x]:
                    for j in range(26):
                            dp[i][j]=max(dp[i][j],dp[x][j])
                    degree[i]-=1
                    if degree[i]==0:
                            q.append(i)
    # print(degree)
    if count!=n:
            print(-1)
    else:
            ans=0
            for i in range(n):
                    ans=max(ans,max(dp[i]))
            print(ans)

beauty(n,m,s,x,y)