CF290-D

D. Fox And Jumping

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 applying i-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 test(s)

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

Note

In first sample test, buying one card is not enough: for example, if you buy a card with length 100, you can't jump to any cell whose index is not a multiple of 100. The best way is to buy first and second card, that will make you be able to jump to any cell.

In the second sample test, even if you buy all cards, you can't jump to any cell whose index is not a multiple of 10, so you should output -1.

给定n个跳跃卡片,卡片中有距离和相应的代价,初始的位置为0,问至少需要多少代价可以跳至任意的位置.

如跳跃10距离的代价为1,那么花费1的代价可以跳至10倍数的任意地方.

要跳至任意距离很容易就想到将所有的卡片组合成能跳跃1距离的"大卡片"

两张卡片能组合成的"大卡片"跳跃距离最小是这两张卡片的最大公约数

对前1至前n张卡片进行遍历,求出能组合成'新距离'的最小代价

状态转移方程 dp[k]=min(dp[k],dp[j]+cost[i]) 其中k=gcd(j,i)

#include <iostream>
using namespace std;
#include <map>
int gcd(long long a,long long b)
{
   return b==0?a:gcd(b,a%b);
}

int main()
{
    map<int,int> dp;
    int n;
    int cost[500];
    int lenth[500];
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>lenth[i];
    }
    for(int i=1;i<=n;i++)
    {
        cin>>cost[i];
    }
    dp[0]=0;
    for(int i=1;i<=n;i++)
    {
        map<int,int>::iterator it=dp.begin();
        for(;it!=dp.end();it++)
        {
            int t=gcd(lenth[i],it->first);
            if(dp.count(t))
            {
                dp[t]=min(dp[t],cost[i]+it->second);
            }
            else
            {
                dp[t]=cost[i]+it->second;
            }
        }
    }
    if(dp.count(1))
        cout<<dp[1]<<endl;
    else
        cout<<"-1"<<endl;
    return 0;
}