p02669 AtCoder Grand Contest 044 - Pay to Win_16100
You start with the number 0 and you want to reach the number N.

You can change the number, paying a certain amount of coins, with the following operations:

* Multiply the number by 2, paying A coins.
* Multiply the number by 3, paying B coins.
* Multiply the number by 5, paying C coins.
* Increase or decrease the number by 1, paying D coins.



You can perform these operations in arbitrary order and an arbitrary number of times.

What is the minimum number of coins you need to reach N?

You have to solve T testcases.

Constraints

* 1 \le T \le 10
* 1 \le N \le 10^{18}
* 1 \le A, B, C, D \le 10^9
* All numbers N, A, B, C, D are integers.

Input

The input is given from Standard Input. The first line of the input is


T


Then, T lines follow describing the T testcases. Each of the T lines has the format


N A B C D


Output

For each testcase, print the answer on Standard Output followed by a newline.

Example

Input

5
11 1 2 4 8
11 1 2 2 8
32 10 8 5 4
29384293847243 454353412 332423423 934923490 1
900000000000000000 332423423 454353412 934923490 987654321


Output

20
19
26
3821859835
23441258666
#include<cstdio>
#include<map>
#include<algorithm>

using namespace std;

long long c[10];

const long long INF=2e18;

map<long long,long long> dp;

long long dfs(long long x)
{
	if(!x)return 0;
	if(dp.count(x))return dp[x];
	long long tmp[6]={x/2,x/2+1,x/3,x/3+1,x/5,x/5+1},t[6]={2,2,3,3,5,5};
	long long ans=INF;for(int i=0;i<6;i++)if(tmp[i]<x)ans=min(ans,dfs(tmp[i])+(long long)min((long double)abs(tmp[i]-x)*c[0],abs(tmp[i]*t[i]-x)*(long double)c[0]+c[t[i]]));
	return dp[x]=ans;
}

int main()
{
	int T=0;scanf("%d",&T);
	while(T--)
	{
		long long x=0;scanf("%lld%lld%lld%lld%lld",&x,&c[2],&c[3],&c[5],&c[0]);
		printf("%lld\n",dfs(x));
		dp.clear();
	}
}