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)

#include <bits/stdc++.h>
using namespace std;

#define rep(i, N) for (int i = 0; i < (N); i++)
#define all(a) (a).begin(), (a).end()
#define pb push_back

using ll = long long;
using i_i = tuple<int, int>;

ll f(ll x, ll a, ll b, ll c, ll d, map<ll, ll>& memo) {
    if (x == 0) return 0;
    if (x == 1) return d;
    if (memo[x] != 0) return memo[x];
    ll ans = LLONG_MAX / 2;
    ans = min(ans, f(x / 2, a, b, c, d, memo) + a + (x - x / 2 * 2) * d);
    ans = min(ans, f((x + 1) / 2, a, b, c, d, memo) + a + ((x + 1) / 2 * 2 - x) * d);
    ans = min(ans, f(x / 3, a, b, c, d, memo) + b + (x - x / 3 * 3) * d);
    ans = min(ans, f((x + 2) / 3, a, b, c, d, memo) + b + ((x + 2) / 3 * 3 - x) * d);
    ans = min(ans, f(x / 5, a, b, c, d, memo) + c + (x - x / 5 * 5) * d);
    ans = min(ans, f((x + 4) / 5, a, b, c, d, memo) + c + ((x + 4) / 5 * 5 - x) * d);
    return memo[x] = ans;
}

int main() {
    int T; cin >> T;
    while (T--) {
        ll x, a, b, c, d;
        cin >> x >> a >> b >> c >> d;
        map<ll, ll> memo;
        cout << f(x, a, b, c, d, memo) << endl;;
    }
}