Codeforces Round #290 (Div. 1)B. Fox And Jumping

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在无限长的带状上，通过购买特定卡片实现从任一位置移动到任一位置的能力，并求解最小成本。每张卡片提供不同长度的跳跃能力与购买成本，挑战在于寻找成本最低的购买组合。

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

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

Input

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

Output

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.

观察后可以发现,如果有a和b两张卡,那么我们可以每次都跳gcd(a, b)

如果要任意格子都能到,说明gcd == 1

于是问题转换成了得到k个数,它们的gcd==1 的最小代价

dp[i] 表示得到i的最小代价

数据有点大,改成用map来转移即可

/*************************************************************************
    > File Name: cf289b.cpp
    > Author: ALex
    > Mail: zchao1995@gmail.com 
    > Created Time: 2015年02月03日 星期二 02时05分37秒
 ************************************************************************/

#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;
map <int, int> dp;
const int N = 330;
int l[N], c[N];

int gcd (int a, int b)
{
	return b == 0 ? a : gcd (b, a % b);
}

int main ()
{
	int n;
	while (~scanf("%d", &n))
	{
		map <int, int> :: iterator it;
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d", &l[i]);
		}
		for (int i = 1; i <= n; ++i)
		{
			scanf("%d", &c[i]);
		}
		dp.clear();
		dp[0] = 0;
		for (int i = 1; i <= n; ++i)
		{
			for (it = dp.begin(); it != dp.end(); ++it)
			{
				int t = gcd (l[i], it -> first);
				if (dp.count (t))
				{
					dp[t] = min (dp[t], it -> second + c[i]);
				}
				else
				{
					dp[t] = it -> second + c[i];
				}
			}
		}
		if (!dp.count (1))
		{
			printf("-1\n");
		}
		else
		{
			printf("%d\n", dp[1]);
		}
	}
	return 0;
}