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.
The first line contains two positive integers $n, m$ ($1 \leq n, m \leq 300\,000$), 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 $x, y$ ($1 \leq x, y \leq n$), 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.
In the first sample, the path with largest value is $1 \to 3 \to 4 \to 5$. The value is $3$ because the letter 'a' appears $3$ times.
5 4
abaca
1 2
1 3
3 4
4 5
3
6 6
xzyabc
1 2
3 1
2 3
5 4
4 3
6 4
-1
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
4
1 1
f
1 1
-1
10 50
ebibwbjihv
1 10
1 2
5 4
1 8
9 7
5 6
1 8
8 7
2 6
5 4
1 9
3 2
8 3
5 6
5 9
2 4
2 7
3 9
1 2
1 7
1 10
3 7
1 8
3 10
8 6
1 7
10 6
1 6
5 8
1 5
2 10
3 9
5 8
8 3
3 7
5 2
1 10
1 4
5 3
3 2
1 2
5 8
10 4
2 10
8 2
1 9
1 8
1 2
3 4
1 8
2
13 37
dwpzcppjmhkmz
2 6
3 6
6 7
6 7
6 7
6 7
6 8
6 8
6 8
6 8
4 6
4 6
5 6
4 6
4 6
6 9
6 9
6 10
6 10
6 10
6 10
4 6
1 6
1 6
10 11
6 11
1 6
6 12
6 12
6 12
6 13
6 13
6 13
6 13
3 6
2 6
2 6
3
5 8
tetqw
2 1
4 4
5 5
5 2
4 5
1 5
1 5
1 1
-1
5 8
reeet
4 3
2 5
4 2
2 4
4 2
5 2
3 3
3 4
-1
from math import *
from collections import *
import math, os, sys, heapq, bisect, random
from functools import lru_cache
from itertools import *
def inp(): return sys.stdin.readline().rstrip("\r\n")
def out(var): sys.stdout.write(str(var))  # for fast output, always take string
def inpu(): return int(inp())
def lis(): return list(map(int, inp().split()))
def stringlis(): return list(map(str, inp().split()))
def sep(): return map(int, inp().split())
def strsep(): return map(str, inp().split())
def fsep(): return map(float, inp().split())
M,M1=1000000007,998244353
#------------------------------------f**k dfs(TLE in python) let's do bfs


def main():
    how_much_noob_I_am = 1
    #how_much_noob_I_am = inpu()
    for case in range(how_much_noob_I_am):
        n,m = sep()
        s= inp()
        d=defaultdict(list)
        indegree=defaultdict(int)
        for i in range(m):
            x,y=sep()
            d[x].append(y)
            indegree[y]+=1
        # def dfs(visited, d, curr, arr, s):
        #     nonlocal ans
        #     ans = max(ans, max(arr))
        #     visited[curr] = 1
        #     for j in d[curr]:
        #         if visited[j] == 1:
        #                 return True
        #         list1 = arr.copy()
        #         list1[ord(s[j - 1]) - 97] += 1
        #         dfs(visited, d, j, list1, s)
        #     visited[curr] = 2
        #     return False
        # for i in range(1,n+1):
        #     if visited[i]==0:
        #         arr=[0]*(26)
        #         arr[ord(s[i - 1]) - 97] += 1
        #         if dfs(visited,d,i,arr,s):
        #             print(-1)
        #             break
        # else:
        #     print(ans)
        q=deque()
        for i in range(1,n+1):
            if indegree[i]==0:
                q.append(i)
        count = 0
        ans = 0
        dp = [[0 for i in range(26)] for i in range(n+1)]
        while count < n and q:
            und = q.popleft()
            count += 1
            dp[und][ord(s[und-1]) - 97] += 1
            for i in d[und]:
                for j in range(26):
                    dp[i][j] = max(dp[i][j], dp[und][j])
                indegree[i] -= 1
                if indegree[i] == 0:
                    q.append(i)
        if count!=n:
            print(-1)
            continue
        for i in range(1,n+1):
            ans = max(ans,max(dp[i]))
        print(ans)
if __name__ == '__main__':
    main()
