Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

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

Sample Output 1

20
19
26
3821859835
23441258666

For the first testcase, a sequence of moves that achieves the minimum cost of 20 is:

Initially x = 0.

Pay 8 to increase by 1 (x = 1).

Pay 1 to multiply by 2 (x = 2).

Pay 1 to multiply by 2 (x = 4).

Pay 2 to multiply by 3 (x = 12).

Pay 8 to decrease by 1 (x = 11).

For the second testcase, a sequence of moves that achieves the minimum cost of 19 is:

Initially x = 0.

Pay 8 to increase by 1 (x = 1).

Pay 1 to multiply by 2 (x = 2).

Pay 2 to multiply by 5 (x = 10).

Pay 8 to increase by 1 (x = 11).

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

/*
これを入れて実行
g++ code.cpp
./a.out
 */
 
#include <iostream>
#include <stdio.h>
#include <vector>
#include <string>
#include <cstring>
#include <queue>
#include <deque>
#include <stack>
#include <algorithm>
#include <utility>
#include <set>
#include <map>
#include <unordered_map>
#include <cmath>
#include <math.h>
#include <tuple>
#include <iomanip>
#include <bitset>
#include <functional>
#include <cassert>

#define all(x) (x).begin(),(x).end()
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
 
using namespace std;
typedef long long ll;
typedef long double ld;
 
int dy4[4] = {-1, 0, +1, 0};
int dx4[4] = {0, +1, 0, -1};
int dy8[8] = {-1, -1, 0, 1, 1, 1, 0, -1};
int dx8[8] = {0, 1, 1, 1, 0, -1, -1, -1};
 
const long long INF = 1LL << 62;
const ll MOD = 1e9 + 7;
 
bool greaterSecond(const pair<int, int>& f, const pair<int, int>& s){
    return f.second > s.second;
}
 
ll gcd(ll a, ll b){
	if (b == 0)return a;
	return gcd(b, a % b);
}
 
ll lcm(ll a, ll b){
    return a / gcd(a, b) * b;
}

ll conbinationMemo[100][100];

void cmemoInit(){
    rep(i, 100){
        rep(j, 100){
            conbinationMemo[i][j] = -1;
        }
    }
}
 
ll nCr(ll n, ll r){
    if(conbinationMemo[n][r] != -1) return conbinationMemo[n][r];
    if(r == 0 || r == n){
        return 1;
    } else if(r == 1){
        return n;
    }
    return conbinationMemo[n][r] = (nCr(n - 1, r) + nCr(n - 1, r - 1));
}
 
ll nPr(ll n, ll r){
    r = n - r;
    ll ret = 1;
    for (ll i = n; i >= r + 1; i--) ret *= i;
    return ret;
}
//-----------------------ここから-----------
map<ll,ll> memo;
ll di[3] = {2, 3, 5};
vector<ll> cost;
ll rec(ll n, ll d, vector<ll> cost){
    if(n == 0){
        return 0;
    }
    if(n == 1){
        return d;
    }

    if(memo.find(n) != memo.end()){
        return memo[n];
    }

    ll res = INF;

    for(int i = 0; i < 3; i++){
        if(n % di[i] == 0){
            res = min(res, rec(n / di[i], d, cost)) + cost[i];
        } else {
            res = min(res, rec((n - (n % di[i])) / di[i], d, cost) + d * (n % di[i]) + cost[i]);
            res = min(res, rec((n + (di[i] - (n % di[i]))) / di[i], d, cost) + d * (di[i] - (n % di[i])) + cost[i]);
        }
    }
    if(n <= INF / d) res = min(res, n * d);

    return memo[n] = res;
}
int main(void){
    ll t;
    cin >> t;
    rep(i, t){
        ll n, a, b, c, d;
        cin >> n >> a >> b >> c >> d;
        cost.clear();
        cost.push_back(a);
        cost.push_back(b);
        cost.push_back(c);
        memo.clear();
        cout << rec(n, d, cost) << endl;
    }
}