CodeForces - 616F：Expensive Strings （后缀自动机）

You are given n strings t_i. Each string has cost c_i.

Let's define the function of string , where p_s, i is the number of occurrences of s in t_i, |s| is the length of the string s. Find the maximal value of function f(s) over all strings.

Note that the string s is not necessarily some string from t.

Input

The first line contains the only integer n (1 ≤ n ≤ 10⁵) — the number of strings in t.

Each of the next n lines contains contains a non-empty string t_i. t_i contains only lowercase English letters.

It is guaranteed that the sum of lengths of all strings in t is not greater than 5·10⁵.

The last line contains n integers c_i ( - 10⁷ ≤ c_i ≤ 10⁷) — the cost of the i-th string.

Output

Print the only integer a — the maximal value of the function f(s) over all strings s. Note one more time that the string s is not necessarily from t.

Examples

Input

2
aa
bb
2 1

Output

4

Input

2
aa
ab
2 1

Output

5

#include<bits/stdc++.h>
#define ll long long
#define rep2(i,a,b) for(int i=a;i>=b;i--)
#define rep(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
const int maxn=;
char c[maxn],s[maxn]; int End[maxn];
struct SAM{
    int ch[maxn][],fa[maxn],maxlen[maxn],cnt,last;
    int a[maxn],b[maxn],num[maxn]; ll val[maxn],ans=;
    void init()
    {
        cnt=; memset(ch[],,sizeof(ch[]));
    }
    int add(int x)
    {
       int np=++cnt,p=last; last=np;
       maxlen[np]=maxlen[p]+;memset(ch[np],,sizeof(ch[np]));
       while(p&&!ch[p][x]) ch[p][x]=np,p=fa[p];
       if(!p) fa[np]=;
       else {
         int q=ch[p][x];
         if(maxlen[q]==maxlen[p]+) fa[np]=q;
         else {
            int nq=++cnt; maxlen[nq]=maxlen[p]+;
            fa[nq]=fa[q]; fa[q]=fa[np]=nq;
            memcpy(ch[nq],ch[q],sizeof(ch[q]));
            while(p&&ch[p][x]==q) ch[p][x]=nq,p=fa[p];
         }
       }
    }
   void sort()
   {
     rep(i,,cnt) a[i]=;
     rep(i,,cnt) a[maxlen[i]]++;
     rep(i,,cnt) a[i]+=a[i-];
     rep(i,,cnt) b[a[maxlen[i]]--]=i;
   }
   void solve(int N)
   {
      sort();
      rep(i,,N) scanf("%d",&num[i]);
      rep(i,,N){
          int now=;
          rep(j,End[i-]+,End[i]){
              now=ch[now][s[j]-'a'];
              val[now]+=num[i];
          }
      }
      rep2(i,cnt,) val[fa[b[i]]]+=val[b[i]];
      rep(i,,cnt) ans=max(ans,val[i]*maxlen[i]);
   }
}T;
int a[maxn];
int main()
{
    int N,L;
    scanf("%d",&N);
    T.init();
    rep(i,,N) {
        scanf("%s",c+);
        L=strlen(c+); T.last=;
        rep(j,,L) T.add(c[j]-'a');
        rep(j,,L) s[End[i-]+j]=c[j];
        End[i]=End[i-]+L;
    }
    T.solve(N);
    printf("%lld\n",T.ans);
    return ;
}

N个串，给个串有一个价值num[i]，现在让你找一个串X，它在第i个字符串出现次数为pi，而且∑len(X)*num[i]*p[i]最大。

思路：后缀自动机，对于每个集合，num之和一样，所以肯定选择最长的一个。