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: Accepted)

#include<bits//stdc++.h>
using namespace std;
typedef unsigned long long int ll;
const ll INF = (1LL << 63);
#define rep(i,n) for(int i = 0; i < n;i++)
int t; ll n, a[10],d;
map<ll, ll> mp;
ll dfs(ll now) {
    if (now == 0) return 0;
    if (now == 1)return d;
    if (mp.count(now)>0) return mp[now]; // すでにあったらリターン
    ll ret = d * now;
    for (int i : {2, 3, 5}) {
        ll next = now / i; // -をするとき
        ret = min(ret, a[i] + (now - next * i) * d + dfs(next));
        next = (now + i - 1) / i; // ＋をするとき
        ret = min(ret, a[i] + (next * i - now) * d + dfs(next));
    }
    return mp[now] = ret; // メモか
}
int main() {
    cin >> t;
    rep(q, t) {
        cin >> n >> a[2] >> a[3] >> a[5] >> d;
        mp.clear();
        cout << dfs(n) << endl;
    }
}