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
#include<bits/stdc++.h>
using namespace std;
#define L long long int
const int M=3e5+10;
int n,m,ans,dp[M][40];
bool mark[M],cyc;
set<int>pth;
vector<int>g[M];
char c[M];
void dfs(int v)
{
  mark[v]=true;
  pth.insert(v);
  for(int i=0;i<g[v].size();i++)
    {
      if(!mark[g[v][i]])
	dfs(g[v][i]);
      else if(pth.find(g[v][i])!=pth.end())
	cyc=true;
      for(char z='a';z<='z';z++)
	dp[v][z-'a']=max(dp[v][z-'a'],dp[g[v][i]][z-'a']);
    }
  dp[v][c[v]-'a']++;
  ans=max(ans,dp[v][c[v]-'a']);
  pth.erase(v);
}
      
int main()
{
  cin>>n>>m;
  scanf("%c",&c[0]);
  for(int i=1;i<=n;i++)
    scanf("%c",&c[i]);
  for(int i=0;i<m;i++)
    {
      int u,v;
      scanf("%d%d",&u,&v);
      g[u].push_back(v);
    }
  for(int i=1;i<=n;i++)
    if(!mark[i])
      dfs(i);
  if(cyc)cout<<-1<<endl;
  else cout<<ans<<endl;
  return 0;
}