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 INF 1e18
#define ll long long
ll n, a, b, c, d;
vector<ll> x;
vector<ll> co;
map<pair<ll, ll>, ll> mp;
ll dp[7];

void gen()
{
    dp[1] = d;
    dp[2] = min(2*d, d+a);
    dp[3] = min(3*d, 2*d+a);
    dp[3] = min(dp[3], d+b);
    dp[3] = min(dp[3], 3*d+c);
    dp[4] = min(4*d, d+2*a);
    dp[4] = min(dp[4], 2*d+b);
    dp[4] = min(dp[4], 2*d+c);
}

ll nows(ll x, ll inc)
{
    if(x == 1) return d;
    ll mins = INF;
    if(x%2 == 0) mins = min(mins, nows(x/2, inc)+a);
    if(x%3 == 0) mins = min(mins, nows(x/3, inc)+b);
    if(x%5 == 0) mins = min(mins, nows(x/5, inc)+c);
    mins = min(mins, nows(x-1, inc)+d);
    if(inc > 0) mins = min(mins, nows(x+1, inc-1)+d);
    return mins;
}

ll mul(ll a, ll b)
{
    if(a >= INF/b) return INF;
    return (a*b);
}

ll calc(ll n, ll wh)
{
    if(n <= 4) return dp[n];
    pair<ll, ll> P = make_pair(n, wh);
    if(mp.find(P) != mp.end()) return mp[P];
    ll mins = INF;
    mins = mul(n, d);
    if(n%x[wh] == 0){
        for(int i=0; i<3; i++){
            mins = min(mins, calc(n/x[wh], i)+co[wh]);
        }
    }
    else {
        ll div = x[wh];
        ll prev = (n/div) * div;
        ll nxt = ((n+div-1)/div)*div;
        ll cost1 = calc(prev, wh) + dp[n-prev];
        ll cost2 = calc(nxt, wh) + dp[nxt-n];
        mins = min(cost1, cost2);
    }
    return mp[P] = mins;
}

int main()
{
    //freopen("in.txt", "r", stdin);
    int cases;
    scanf("%d", &cases);
    while(cases--){
        scanf("%lld %lld %lld %lld %lld", &n, &a, &b, &c, &d);
        gen();
        x = {2, 3, 5};
        co = {a, b, c};
        ll ans = INF;
        for(int i=0; i<3; i++) ans = min(ans, calc(n, i));
        cout << ans << endl;
        mp.clear();
    }
}