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 long long ll;
typedef pair<long long, long long> P;
struct edge{long long to,cost;};
const int inf = 1 << 27;
const long long INF = 1LL << 60;
const int COMBMAX = 1001001;
const long long MOD = 1000000007;
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define eachdo(v, e) for(const auto& e : (v))
#define all(v) (v).begin(), (v).end()
#define lower_index(v, e) (long long)distance((v).begin(), lower_bound((v).begin(), (v).end(), e))
#define upper_index(v, e) (long long)distance((v).begin(), upper_bound((v).begin(), (v).end(), e))
long long mpow(long long a, long long n, long long mod = MOD){long long res = 1; while(n > 0){if(n & 1)res = res * a % mod; a = a * a % mod; n >>= 1;} return res;}
void yn(bool j){cout << (j ? "Yes" : "No") << endl; return;}
template<class Head> void pt(Head&& head){cout << head << endl; return;}
template<class Head, class... Tail> void pt(Head&& head, Tail&&... tail){cout << head << " "; pt(forward<Tail>(tail)...);}
template<class T> void debug(T v){rep(i, v.size()) cout << v[i] << " " ; cout << endl;}
template<class T> void debug2(T v){rep(i, v.size()){rep(j, v[i].size()) cout << v[i][j] << " " ; cout << endl;}}
template<class T1, class T2> long long bcount(T1 v, T2 a){return upper_index(v, a) - lower_index(v, a);} 
template<class T1, class T2> inline bool chmin(T1 &a, T2 b){if(a > b){a = b; return true;} return false;}
template<class T1, class T2> inline bool chmax(T1 &a, T2 b){if(a < b){a = b; return true;} return false;}
ll N, A, B, C, D;
ll width = 6;
unordered_map<ll, ll> dp;

ll rec(ll now){
    if (dp.count(now)) return dp[now];
    if (now == 0) return 0;
    ll ret = INF;
    if (INF / D > now) ret = D * now;
    for (ll i = now - width; i < now + width - 3; i++){
        if (i < 0) continue;
        ll d = abs(now - i);
        if ((i % 2) == 0 && i / 2 != now) chmin(ret, rec(i / 2) + A + d * D);
        if ((i % 3) == 0 && i / 3 != now) chmin(ret, rec(i / 3) + B + d * D);
        if ((i % 5) == 0 && i / 5 != now) chmin(ret, rec(i / 5) + C + d * D);
    }
    // pt(now);
    dp[now] = ret;
    return ret;
}

void solve(){
    cin >> N >> A >> B >> C >> D;
    dp.clear();
    pt(rec(N));
}



int main(){
    ll T; cin >> T;
    while(T--){
        solve();
    }
}