Codeforces Round #290 (Div. 2)-D. Fox And Jumping

原題鏈接

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 li and cost ci. If she pays ci dollars then she can apply i-th card. After applying i-th card she becomes able to make jumps of length li, i. e. from cell x to cell (x - li) or cell (x + li).

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 li (1 ≤ li ≤ 109), the jump lengths of cards.

The third line contains n numbers ci (1 ≤ ci ≤ 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.

Examples

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

只有找到一組數他們加減之後爲1即可，那麼這組數的最大公約數必須爲1

#include <bits/stdc++.h>
#define maxn 100005
#define MOD 1000000007
using namespace std;
typedef long long ll;

int k1[305], k2[305];
map<int, int> m;
int gcd(int a, int b){
	return b ? gcd(b, a % b) : a;
}
int main(){
//	freopen("in.txt", "r", stdin);
	int n;
	scanf("%d", &n);
	for(int i = 0; i < n; i++)
	 scanf("%d", k1+i);
	for(int i = 0; i < n; i++)
	 scanf("%d", k2+i);
	map<int, int> ::iterator iter;
	for(int i = 0; i < n; i++){
		if(m.count(k1[i]))
		  m[k1[i]] = min(m[k1[i]], k2[i]);
		else
		 m[k1[i]] = k2[i];
		iter = m.begin();
		for(; iter != m.end(); ++iter){
			int a = gcd(iter->first, k1[i]);
			if(m.count(a))
			 m[a] = min(m[a], iter->second+k2[i]);
			else
			 m[a] = iter->second + k2[i];
		}
	}
	if(m.count(1) == 0)
	 puts("-1");
	else
	 printf("%d\n", m[1]);
	return 0;
}