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)

#ifdef LOCAL
//#define _GLIBCXX_DEBUG
#endif
//#pragma GCC target("avx512f,avx512dq,avx512cd,avx512bw,avx512vl")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef pair<ll, ll> P;
typedef pair<int, int> Pi;
typedef vector<ll> Vec;
typedef vector<int> Vi;
typedef vector<string> Vs;
typedef vector<P> VP;
typedef vector<vector<ll>> VV;
typedef vector<vector<int>> VVi;
typedef vector<vector<vector<ll>>> VVV;
typedef vector<vector<vector<vector<ll>>>> VVVV;

#define REP(i, a, b) for(ll i=(a); i<(b); i++)
#define PER(i, a, b) for(ll i=(a); i>=(b); i--)
#define rep(i, n) REP(i, 0, n)
#define per(i, n) PER(i, n, 0)
const ll INF = 1e16+18;
const ll MOD = 1000000007;
#define Yes(n) cout << ((n) ? "Yes" : "No") << endl;
#define YES(n) cout << ((n) ? "YES" : "NO") << endl;
#define ALL(v) v.begin(), v.end()
#define rALL(v) v.rbegin(), v.rend()
#define pb(x) push_back(x)
#define mp(a, b) make_pair(a,b)
#define Each(a, b) for(auto &a :b)
#define REPM(i, mp) for (auto i = mp.begin(); i != mp.end(); ++i)
#ifdef LOCAL
#define dbg(x_) cerr << #x_ << ":" << x_ << endl;
#define dbgmap(mp) cerr << #mp << ":"<<endl; for (auto i = mp.begin(); i != mp.end(); ++i) { cerr << i->first <<":"<<i->second << endl;}
#define dbgarr(n, m, arr) rep(i,n){rep(j,m){cerr<<arr[i][j]<<" ";}cerr<<endl;}
#define dbgdp(n, arr) rep(i,n){cerr<<arr[i]<<" ";}cerr<<endl;
#define dbgmint(n, arr) rep(i,n){cerr<<arr[i].x<<" ";}cerr<<endl;
#define dbgarrmint(n, m, arr) rep(i,n){rep(j,m){cerr<<arr[i][j].x<<" ";}cerr<<endl;}
#else
#define dbg(...)
#define dbgmap(...)
#define dbgarr(...)
#define dbgdp(...)
#define dbgmint(...)
#define dbgarrmint(...)
#endif
#define out(a) cout<<a<<endl
#define out2(a, b) cout<<a<<" "<<b<<endl
#define vout(v) rep(i,v.size()){cout<<v[i]<<" ";}cout<<endl
#define Uniq(v) v.erase(unique(v.begin(), v.end()), v.end())
#define fi first
#define se second

template<class T>
bool chmax(T &a, const T &b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

template<class T>
bool chmin(T &a, const T &b) {
    if(b < a) {
        a = b;
        return true;
    }
    return false;
}

template<typename T1, typename T2>
ostream &operator<<(ostream &s, const pair<T1, T2> &p) { return s<<"("<<p.first<<", "<<p.second<<")"; }

template<typename T>
istream &operator>>(istream &i, vector<T> &v) {
    rep(j, v.size())i>>v[j];
    return i;
}

// vector
template<typename T>
ostream &operator<<(ostream &s, const vector<T> &v) {
    int len = v.size();
    for(int i = 0; i < len; ++i) {
        s<<v[i];
        if(i < len-1) s<<"	";
    }
    return s;
}

// 2 dimentional vector
template<typename T>
ostream &operator<<(ostream &s, const vector<vector<T> > &vv) {
    int len = vv.size();
    for(int i = 0; i < len; ++i) {
        s<<vv[i]<<endl;
    }
    return s;
}


//mint
struct mint {
    ll x; // typedef long long ll;
    mint(ll x = 0) : x((x%MOD+MOD)%MOD) {}

    mint &operator+=(const mint a) {
        if((x += a.x) >= MOD) x -= MOD;
        return *this;
    }

    mint &operator-=(const mint a) {
        if((x += MOD-a.x) >= MOD) x -= MOD;
        return *this;
    }

    mint &operator*=(const mint a) {
        (x *= a.x) %= MOD;
        return *this;
    }

    mint operator+(const mint a) const {
        mint res(*this);
        return res += a;
    }

    mint operator-(const mint a) const {
        mint res(*this);
        return res -= a;
    }

    mint operator*(const mint a) const {
        mint res(*this);
        return res *= a;
    }

    mint pow(ll t) const {
        if(!t) return 1;
        mint a = pow(t>>1);
        a *= a;
        if(t & 1) a *= *this;
        return a;
    }

    // for prime MOD
    mint inv() const {
        return pow(MOD-2);
    }

    mint &operator/=(const mint a) {
        return (*this) *= a.inv();
    }

    mint operator/(const mint a) const {
        mint res(*this);
        return res /= a;
    }
};

typedef vector<mint> Vmint;

struct combination {
    vector<mint> fact, ifact;

    combination(int n) : fact(n+1), ifact(n+1) {
        assert(n < MOD);
        fact[0] = 1;
        for(int i = 1; i <= n; ++i) fact[i] = fact[i-1]*i;
        ifact[n] = fact[n].inv();
        for(int i = n; i >= 1; --i) ifact[i-1] = ifact[i]*i;
    }

    mint operator()(int n, int k) {
        if(k < 0 || k > n) return 0;
        return fact[n]*ifact[k]*ifact[n-k];
    }
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout<<std::setprecision(10);

    ll T;
    cin>>T;
    while (T--) {
        ll n, a, b, c, d;
        cin>>n>>a>>b>>c>>d;

        map<ll, ll> memo;
        auto dfs = [&](auto dfs, ll x)->ll {
            if(memo.count(x) > 0) {
                return memo[x];
            }
            if(x == 0) {
                return memo[x] = 0;
            }
            if(x == 1) {
                return memo[x] = d;
            }

            ll res = INF;
            if(x < 1e18/d) res = x*d;

            ll da = x%2;
            ll db = x%3;
            ll dc = x%5;

            ll c1 = (x-dc+5)/5;
            ll c2 = (x-dc)/5;

            if(dc > 0) {
                chmin(res, (5-dc)*d + c +dfs(dfs, c1));
            }
            chmin(res, dc*d + c +dfs(dfs, c2));

            ll b1 = (x-db+3)/3;
            ll b2 = (x-db)/3;

            if(db > 0) {
                chmin(res, (3-db)*d + b + dfs(dfs, b1));
            }
            chmin(res, db*d + b + dfs(dfs, b2));

            ll a1 = (x-da+2)/2;
            ll a2 = (x-da)/2;

            if(da > 0) {
                chmin(res, (2-da)*d + a + dfs(dfs, a1));
            }
            chmin(res, da*d + a + dfs(dfs, a2));

            return memo[x] = res;
        };

        out(dfs(dfs, n));

    }
    return 0;
}