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;
 
typedef long long ll;
#define pa pair<int, int>
#define pal pair<long long, long long>
#define pali pair<long long, int>
#define pad pair<double, double>
#define pb push_back
#define mp make_pair
#define COUT(v)                              \
    for (int64_t i = 0; i < (v).size(); ++i) \
    {                                        \
        cout << v.at(i) << endl;             \
    }
#define REP(i, n) for (int64_t i = 0; i < n; ++i)
#define FOR(i, r, n) for (int64_t i = (r); i < n; ++i)
#define VIN(v)                               \
    for (int64_t i = 0; i < (v).size(); ++i) \
    {                                        \
        cin >> (v).at(i);                    \
    }
 
typedef vector<bool> bvec;
typedef vector<int> ivec;
typedef vector<long long> lvec;
typedef vector<double> dvec;
typedef vector<pa> pavec;
typedef vector<pali> palivec;
typedef vector<pal> palvec;
 
typedef vector<vector<bool> > bmat;
typedef vector<vector<int> > imat;
typedef vector<vector<long long> > lmat;
 
typedef vector<string> svec;
typedef vector<vector<string> > smat;
const ll infll = (1LL << 60) - 1;
const int inf = (1 << 30) - 1;
const ll MOD = 1000000007;
 
ll gcd(ll x, ll y)
{
    ll r = x % y;
    if (r == 0)
        return y;
    else
        return gcd(y, r);
}
 
ll lcm(ll x, ll y)
{
    return x * y / gcd(x, y);
}
 
lvec mfactor(ll n)
{
    bvec ip(n, true);
    lvec mf(n, -1);
    ip[0] = false;
    ip[1] = false;
    mf[0] = 0;
    mf[1] = 1;
    REP(i, n)
    {
        if (ip[i])
        {
            mf[i] = i;
            for (ll j = i * i; j < n; j += i)
            {
                ip[j] = false;
                if (mf[j] == -1)
                    mf[j] = i;
            }
        }
    }
    return mf;
}
 
palivec get_prime(ll n, const lvec &mf)
{
    palivec plist;
    while (n != 1)
    {
        int cnt = 0;
        ll m = mf[n];
        while (mf[n] == m)
        {
            cnt++;
            n /= m;
        }
        plist.pb(pali(m, cnt));
    }
    return plist;
}
 
void COMinit(int m, lvec &fac, lvec &finv)
{
    lvec inv(m);
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < m; i++)
    {
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
 
    }
}
 
ll pow_mod(ll a, ll n)
{
    ll x = 1;
    while (n > 0)
    {
        if (n & 1)
        {
            x = x * a % MOD;
        }
        a = a * a % MOD;
        n >>= 1;
    }
    return x;
}

ll pow_mod2(ll a, ll n)
{
    ll x = 1;
    while (n > 0)
    {
        if (n & 1)
        {
            x = x * a;
        }
        a = a * a;
        n >>= 1;
    }
    return x;
}
 
 
ll COM(int n, int k, const lvec &fac, const lvec &finv)
{
    if (n < k)
        return 0;
    if (n < 0 || k < 0)
        return 0;
    return (fac[n] * (finv[k] * finv[n - k] % MOD)) % MOD;
}
 
ll modinv(ll a, ll m)
{
    ll b = m, u = 1, v = 0;
    while (b)
    {
        ll t = a / b;
        a -= t * b;
        swap(a, b);
        u -= t * v;
        swap(u, v);
    }
    u %= m;
    if (u < 0)
        u += m;
    return u;
}

// union by size + path having
class UnionFind {
public:
    vector <ll> par; // 各元の親を表す配列
    vector <ll> siz; // 素集合のサイズを表す配列(1 で初期化)

    // Constructor
    UnionFind(ll sz_): par(sz_), siz(sz_, 1LL) {
        for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
    }
    void init(ll sz_) {
        par.resize(sz_);
        siz.assign(sz_, 1LL);  // resize だとなぜか初期化されなかった
        for (ll i = 0; i < sz_; ++i) par[i] = i; // 初期では親は自分自身
    }

    // Member Function
    // Find
    ll root(ll x) { // 根の検索
        while (par[x] != x) {
            x = par[x] = par[par[x]]; // x の親の親を x の親とする
        }
        return x;
    }

    // Union(Unite, Merge)
    bool merge(ll x, ll y) {
        x = root(x);
        y = root(y);
        if (x == y) return false;
        // merge technique（データ構造をマージするテク．小を大にくっつける）
        if (siz[x] < siz[y]) swap(x, y);
        siz[x] += siz[y];
        par[y] = x;
        return true;
    }

    bool issame(ll x, ll y) { // 連結判定
        return root(x) == root(y);
    }

    ll size(ll x) { // 素集合のサイズ
        return siz[root(x)];
    }
};

ll solve(ll a, ll b, ll c, ll d, ll n){
    map<ll, ll> m;
    m[0] = 0;
    m[1] = d;
    m[2] = m[1] + min(d, a);
    function<ll(ll)> dfs = [&](ll x){
        
        if(m.count(x)) return m[x];    
        ll wa = min(dfs(x/2) + d*(x - 2*(x/2)) + a, dfs(x/2 + 1) + d*((x/2+1)*2 - x) + a);
        ll wb = min(dfs(x/3) + d*(x - 3*(x/3)) + b, dfs(x/3 + 1) + d*((x/3+1)*3 - x) + b);
        ll wc = min(dfs(x/5) + d*(x - 5*(x/5)) + c, dfs(x/5 + 1) + d*((x/5+1)*5 - x) + c);
        m[x] = min(wa, wb);
        m[x] = min(m[x], wc);
        ll xd = x*d;
        if(xd/d == x) m[x] = min(m[x], x*d);
        return m[x];
    };
    dfs(n);
    
    return m[n];
}

int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll t;
    cin >> t;
    lvec avec(t);
    for (int i=0; i<t; i++){
        ll n, a, b, c, d;
        cin >> n >> a >> b >> c >> d;
        avec[i] = solve(a, b, c, d, n);
    }
    for (auto v: avec){
        cout << v << endl;
    }

    
    
    }