Codeforces Round #460 (Div. 2): D. Substring（DAG+DP+判环）

D. Substring

time limit per test

3 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

题意：给一个由字母代替节点的有向图，一条路径的值为出现次数最多的字母出现次数，求出路径的最大值（如果无穷大输出-1）

思路：有环就是-1，接下来就是DAG，暴力26个字母，对于当前字母x，将所有为x的节点权值设为1，其它节点权值设为0，就是一个非常简单的DP了。

代码：

 //#include "bits/stdc++.h"
 #include "cstdio"
 #include "map"
 #include "set"
 #include "cmath"
 #include "queue"
 #include "vector"
 #include "string"
 #include "cstring"
 #include "time.h"
 #include "iostream"
 #include "stdlib.h"
 #include "algorithm"
 #define db double
 #define ll long long
 #define vec vector<ll>
 #define Mt  vector<vec>
 #define ci(x) scanf("%d",&x)
 #define cd(x) scanf("%lf",&x)
 #define cl(x) scanf("%lld",&x)
 #define pi(x) printf("%d\n",x)
 #define pd(x) printf("%f\n",x)
 #define pl(x) printf("%lld\n",x)
 #define inf 0x3f3f3f3f
 #define rep(i, x, y) for(int i=x;i<=y;i++)
 const int N   = 3e5 + ;
 const int mod = 1e9 + ;
 const int MOD = mod - ;
 const db  eps = 1e-;
 const db  PI  = acos(-1.0);
 using namespace std;
 vector<int> g[N];
 queue<int> q;
 char s[N];
 int in[N],deg[N],vis[N],f[N],val[N];
 int n,m,ans;
 bool cal()//判环
 {
     int cnt=;
     for(int i=;i<=n;i++){
         if(!in[i]) q.push(i),cnt++;
         vis[i]=;
     }
     while(q.size()){
         int v=q.front();
         q.pop();
         for(int i=;i<g[v].size();i++){
             int vv=g[v][i];
             in[vv]--;
             if(!in[vv]) vis[vv]=,q.push(vv),cnt++;
         }
     }
     return cnt==n;
 }
 void dfs(int u)//DP（DAG）
 {
     f[u]=val[u];
     vis[u]=;
     for(int i=;i<g[u].size();i++){
         int v=g[u][i];
         if(!vis[v]) dfs(v);
         f[u]=max(f[u],f[v]+val[u]);
     }
     ans=max(f[u],ans);
 }
 int main()
 {
     ci(n),ci(m);
     scanf("%s",s+);
     for(int i=;i<m;i++){
         int x,y;
         ci(x),ci(y);
         g[x].push_back(y);
         in[y]++,deg[y]++;
     }
     if(!cal()) puts("-1");
     else
     {
         for(int i=;i<;i++){
             memset(vis,, sizeof(vis));
             memset(f,, sizeof(f));
             for(int j=;j<=n;j++){//节点赋值
                 if(s[j]=='a'+i) val[j]=;
                 else val[j]=;
             }
             for(int j=;j<=n;j++){
                 if(!deg[j]) dfs(j);
             }
         }
         pi(ans);
     }
     return ;
 }