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 collections import defaultdict
class Graph:
 
  def __init__(self, n, s):
    self.n=n
    self.s=s
    self.adjMatrix={}
    self.costArray=defaultdict(list)
    for i in range(n):
      self.costArray[i]=[]
      for j in range(26):
        self.costArray[i].append(0)
 
  def addEdge(self, x, y):
    x-=1
    y-=1
    if x not in self.adjMatrix:
      self.adjMatrix[x]=set()
    self.adjMatrix[x].add(y)
 
  def hasCycle(self, x, isVisitedWhole, isVisitedInPath):
    if isVisitedInPath[x] is True:
      return True
    if isVisitedWhole[x] is True:
      return False
    isVisitedWhole[x]=isVisitedInPath[x]=True
    if x in self.adjMatrix:
      for i in self.adjMatrix[x]:
        if self.hasCycle(i, isVisitedWhole, isVisitedInPath):
          return True
    isVisitedInPath[x]=False
    return False
 
  def dfs(self, x, isVisited):
    isVisited[x]=True
    if x in self.adjMatrix:
      for i in self.adjMatrix[x]:
        if isVisited[i] is False:
          self.dfs(i, isVisited)
        for j in range(26):
          self.costArray[x][j]=max(self.costArray[x][j], self.costArray[i][j]);
    self.costArray[x][ord(self.s[x])-ord('a')]+=1
 
  def maximumCost(self):
    isVisitedWhole={}
    isVisitedInPath={}
    for i in range(self.n):
      isVisitedWhole[i]=False
      isVisitedInPath[i]=False
    for i in range(self.n):
      if isVisitedWhole[i] is False and self.hasCycle(i, isVisitedWhole, isVisitedInPath):
        return -1
    isVisited={}
    for i in range(self.n):
      isVisited[i]=False
    for i in range(self.n):
      if isVisited[i] is False:
        self.dfs(i, isVisited)
    costMax=1
    for i in range(self.n):
      for j in range(26):
        costMax=max(costMax, self.costArray[i][j])
    return costMax
 
 
n, m=map(int, input().split())
s=input()
graph=Graph(n, s)
for i in range(m):
  x, y=map(int, input().split())
  graph.addEdge(x, y)
print(graph.maximumCost())