D – Fox And Jumping

　　　　　　　　　　　　　　　　　　　　　　Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 512B

Description

Fox Ciel is playing a game. In this game there is an infinite long tape with cells indexed by integers (positive, negative and zero). At the beginning she is standing at the cell 0.

There are also n cards, each card has 2 attributes: length l_i and cost c_i. If she pays c_i dollars then she can apply i-th card. After applyingi-th card she becomes able to make jumps of length l_i, i. e. from cell x to cell (x - l_i) or cell (x + l_i).

She wants to be able to jump to any cell on the tape (possibly, visiting some intermediate cells). For achieving this goal, she wants to buy some cards, paying as little money as possible.

If this is possible, calculate the minimal cost.

Input

The first line contains an integer n (1 ≤ n ≤ 300), number of cards.

The second line contains n numbers l_i (1 ≤ l_i ≤ 10⁹), the jump lengths of cards.

The third line contains n numbers c_i (1 ≤ c_i ≤ 10⁵), the costs of cards.

Output

If it is impossible to buy some cards and become able to jump to any cell, output -1. Otherwise output the minimal cost of buying such set of cards.

Sample Input

Input

3
100 99 9900
1 1 1

Output

2

Input

5
10 20 30 40 50
1 1 1 1 1

Output

-1

Input

7
15015 10010 6006 4290 2730 2310 1
1 1 1 1 1 1 10

Output

6

Input

8
4264 4921 6321 6984 2316 8432 6120 1026
4264 4921 6321 6984 2316 8432 6120 1026

Output

7237区间dp

 //2016.8.6
 #include<iostream>
 #include<cstdio>
 #include<map>
 #include<algorithm> using namespace std; int l[], c[];
 map<int, int>dp1, dp2; int gcd(int a, int b)
 {
     return b==?a:gcd(b, a%b);
 } int main()
 {
     int n;
     while(cin>>n)
     {
         dp1.clear();
         for(int i = ; i <= n; i++)
           scanf("%d", &l[i]);
         for(int i = ; i <= n; i++)
           scanf("%d", &c[i]);
         dp1[] = ;
         for(int i = ; i <= n; i++)
         {
             map<int, int>::iterator it;
             for(it = dp1.begin(); it != dp1.end(); it++)
             {
                 int g = gcd(it->first, l[i]);
                 if(dp1.count(g))
                   dp1[g] = min(it->second+c[i], dp1[g]);
                 else dp1[g] = it->second+c[i];
             }
         }
         if(dp1.count())cout<<dp1[]<<endl;
         else cout<<-<<endl;
     }     return ;
 }