没想象中难的substring 有向图判环 拓补· dfs

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 3because the letter 'a' appears 3 times.

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#include <cstdio>
#include <cmath>
#include <vector>
#include <queue>
#include <cstring>
#include <algorithm>
using namespace std;
#define maxn 300005
#define mst(a,b) memset((a),(b),sizeof(a))
int n,m;
int degree[maxn];
char s[maxn];
int dp[maxn];
vector<int>vec[maxn];
vector<int>mp;

bool check()
{
    mp.clear();
    queue<int>q;
    for(int i=1;i<=n;i++)
    {
        if(degree[i]==0) q.push(i);
    }
    while(q.size())
    {
        int u=q.front();
        q.pop();
        mp.push_back(u);
        for(int i=0;i<vec[u].size();i++)
        {
            degree[vec[u][i]]--;
            if(degree[vec[u][i]]==0) q.push(vec[u][i]);
        }
    }
    return mp.size()==n;
}

int solve(int x)
{
    mst(dp,0);
    char c=x+'a';
    int cnt=0;
    for(int i=mp.size()-1;i>=0;i--)
    {
        int u=mp[i];
        for(int j=0;j<vec[u].size();j++)
        {
            dp[u]=max(dp[u],dp[vec[u][j]]);
        }
        if(s[u-1]==c) dp[u]++;
        cnt=max(cnt,dp[u]);
    }
    return cnt;
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++) vec[i].clear();
        scanf("%s",s);
        mst(degree,0);
        for(int i=0;i<m;i++)
        {
            int x,y;
            scanf("%d%d",&x,&y);
            vec[x].push_back(y);
            degree[y]++;
        }
        if(!check())
        {
            puts("-1");
            continue;
        }
        int ans=0;
        for(int i=0;i<26;i++)
        {
            ans=max(ans,solve(i));
        }
        printf("%d\n",ans);
    }
}