Problem:
Score: 300 points

Problem Statement

In 2028 and after a continuous growth, AtCoder Inc. finally built an empire with six cities (City 1, 2, 3, 4, 5, 6)!

There are five means of transport in this empire:

Train: travels from City 1 to 2 in one minute. A train can occupy at most A people.

Bus: travels from City 2 to 3 in one minute. A bus can occupy at most B people.

Taxi: travels from City 3 to 4 in one minute. A taxi can occupy at most C people.

Airplane: travels from City 4 to 5 in one minute. An airplane can occupy at most D people.

Ship: travels from City 5 to 6 in one minute. A ship can occupy at most E people.

For each of them, one vehicle leaves the city at each integer time (time 0, 1, 2, ...).

There is a group of N people at City 1, and they all want to go to City 6.

At least how long does it take for all of them to reach there?
You can ignore the time needed to transfer.

Constraints

1 \leq N, A, B, C, D, E \leq 10^{15}

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A
B
C
D
E

Output

Print the minimum time required for all of the people to reach City 6, in minutes.

Sample Input 1

5
3
2
4
3
5

Sample Output 1

7

One possible way to travel is as follows.
First, there are N = 5 people at City 1, as shown in the following image:

In the first minute, three people travels from City 1 to City 2 by train. Note that a train can only occupy at most three people.

In the second minute, the remaining two people travels from City 1 to City 2 by train, and two of the three people who were already at City 2 travels to City 3 by bus. Note that a bus can only occupy at most two people.

In the third minute, two people travels from City 2 to City 3 by train, and another two people travels from City 3 to City 4 by taxi.

From then on, if they continue traveling without stopping until they reach City 6, all of them can reach there in seven minutes.

There is no way for them to reach City 6 in 6 minutes or less.

Sample Input 2

10
123
123
123
123
123

Sample Output 2

5

All kinds of vehicles can occupy N = 10 people at a time.
Thus, if they continue traveling without stopping until they reach City 6, all of them can reach there in five minutes.

Sample Input 3

10000000007
2
3
5
7
11

Sample Output 3

5000000008

Note that the input or output may not fit into a 32-bit integer type.

Solution in C++14 (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i,n) for (int i = 0; i< (n); ++i)
const long long INF = 1LL<<58;

// aを常に小さい値にする　大きい値にする
template<class T> bool chmin(T &a, T b){
    if(a>b) {
        a = b;
        return true;
    }
    return false;

}
template<class T> void chmax(T &a, T b){ if(a<b) a = b; }

// 二分探索 前方のindexを返すもの（自分自身を含む） 前方: -INF側のこと
template<class T> int former(const vector<T> &v, T x){
    return upper_bound(v.begin(), v.end(), x) - v.begin() -1; // upper_bound : keyより大きい要素の内の一番左側のイテレータを返す
}
// 二分探索 後方のindexを返すもの（自分自身を含む）後方: INF側のこと
template<class T> int latter(const vector<T> &v, T x){
    return lower_bound(v.begin(), v.end(), x) - v.begin(); // lower_bound : key以上の要素の内の一番左側のイテレータを返す
}

// 最大公約数を求める GCD
int GCD(int a, int b) { return b ? GCD(b, a%b) : a; }


/*
 *
 * */

int main() {

    ll N, A,B,C,D,E;
    cin >> N >> A >> B >> C >> D >> E;

    vector<ll> status(6, 0);
    status[0] = N;

    ll count = 0;
    while(true){
        count++;

        if(status[4]-E >= 0){
            status[5] += E;
            status[4] -= E;
        }else{
            status[5] += status[4];
            status[4] = 0;
        }


        if(status[3]-D >= 0){
            status[4] += D;
            status[3] -= D;
        }else{
            status[4] += status[3];
            status[3] = 0;
        }


        if(status[2]-C >= 0){
            status[3] += C;
            status[2] -= C;
        }else{
            status[3] += status[2];
            status[2] = 0;
        }


        if(status[1]-B >= 0){
            status[2] += B;
            status[1] -= B;
        }else{
            status[2] += status[1];
            status[1] = 0;
        }

        if(status[0]-A >= 0){
            status[1] += A;
            status[0] -= A;
        }else{
            status[1] += status[0];
            status[0] = 0;
        }

        if(status[5] == N){
            break;
        }
    }

    cout << count << endl;
}