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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.math.BigInteger;
import java.util.*;
 
 
public class Main {
 
	static long MOD = 1_000_000_007;
	static int INF = 1_000_000_007;
	//static long INF = Long.MAX_VALUE;
	//二項係数関連ここから-----
	static boolean isPreprocessing = false;	//二項係数を計算するための前処理を行っているかどうかのフラグ
	static int MAX = 666667;				//MAXまで階乗やそれらの逆元のテーブルを保持する。
	static long[] fac;	//fac[i] := i! % MOD
	static long[] inv;	//inv[i] := mod. MODにおける i の逆元
	static long[] finv;	//finv[i] := mod. MODにおける i! の逆元(invの累積"積")
	//二項係数関連おわり------
	
	static int[] roots;		//Union-find木用
	static int[] ans;
	//static List<ArrayList<Integer>> G;	//グラフ用

	static List<List<E>> G = new ArrayList<>();	//グラフ用
	static List<E> Edges = new ArrayList<>();	//グラフ用
	static int V;								//グラフ用頂点の数
	static int E;								//グラフ用辺の数
	static long[] X = new long[1];				//拡張ユークリッドの互除法用
	static long[] Y = new long[1];				//拡張ユークリッドの互除法用
	//旧入力
	//Scanner sc = new Scanner(System.in);
	//int N = Integer.parseInt(sc.next());
	static int[] dx = {1,-1,0,0};
	static int[] dy = {0,0,1,-1};
	static int[] dx8 = {1,-1,0,0,1,1,-1,-1};
	static int[] dy8 = {0,0,1,-1,1,-1,1,-1};
	
	static int[] R; 
	public static void main(String[] args) throws Exception {
		FastScanner scan = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		//int N = scan.nextInt();
		//long L = scan.nextLong();
		//char[] A = scan.next().toCharArray();
		
		long N = scan.nextLong();
		int ind = 1;
		long A = scan.nextLong();
		long min = A;
		
		long B = scan.nextLong();
		if(B<min){
			min = B;
			ind = 2;
		}
		long C = scan.nextLong();
		if(C<min){
			min = C;
			ind = 3;
		}
		long D = scan.nextLong();
		if(D<min){
			min = D;
			ind = 4;
		}
		long E = scan.nextLong();
		if(E<min){
			min = E;
			ind = 5;
		}
		
		long ans = 0;
		if(min == A){
			ans = N/min + 6 - ind;
		}else{
			ans = (N+min-1)/min + 6 - ind;
		}
		
		
		
		
		
		out.println(ans);	
	
		out.flush();
	}

	//i_C_jをMODで割ったあまりを求める（計算量最悪）
	static long com(long i, long j){
		long rtn = 1L;
		for(long z=i; z>(i-j); z--){
			rtn *= z;
			rtn %= MOD;
		}
		for(long z=1; z<=j; z++){
			rtn *= modinv(z, MOD);
			rtn %= MOD;
		}
		return rtn;
	}

	//a_i≧kとなるような最小のiを求める(0≦i≦N-1)
	//存在しなかったら-1を返す
	static int lower_bound(long[] A, long k){
		int N = A.length;
		int lb = -1;
		int ub = N;
		while(ub-lb > 1){
			int mid = (lb + ub) / 2;
			if(A[mid] >= k){
				ub = mid;
			}else{
				lb = mid;
			}
		}
		return ub;
	}

	//k≧a_iとなるような最大のiを求める(0≦i≦N-1)
	//存在しなかったらa_N-1を返す
	static int upper_bound(long[] A, long k){
		int N = A.length;
		int lb = -1;
		int ub = N;
		while(ub-lb > 1){
			int mid = (lb + ub) / 2;
			if(A[mid] <= k){
				lb = mid;
			}else{
				ub = mid;
			}
		}
		if(lb == N) lb = N-1;
		return lb;
	}
	
	//aの逆元a^-1を求める
	//条件：aとmodが互いに素であること(aがmodの倍数でないこと)
	static long modinv(long a, long mod){
		long b = mod;
		long u = 1;
		long v = 0;
		while(b>0){
			long t = a/b;
			a -= t*b;
			long tmp = b;
			b = a;
			a = tmp;
			u -= t*v;
			tmp = v;
			v = u;
			u = tmp;
		}
		u %= mod;
		if(u<0) u+=mod;
		return u;
	}
	
	//繰り返し二乗法(x^n)
	static long pow(long x, long n){
		long sum = 1;
		while(n>0){
			if((n&1) == 1 ){
				sum = sum * x;
			}
			x = x * x;
			n >>= 1;
		}
		return sum;
	}
	
	//繰り返し二乗法(x^n % mod)
	static long modPow(long x, long n, long mod){
		long sum = 1;
		while(n>0){
			if((n&1) == 1 ){
				sum = sum * x % mod;
			}
			x = (x * x) % mod;
			n >>= 1;
		}
		return sum;
	}
	
	//オイラーのφ関数（nに対して、nと互いに素である1以上n以下の自然数の個数）
	static int phi(int n){
		List<long[]> primes = factorization(n);//素因数分解
		int rtn = n;
		for(long[] prime : primes){
			int p = (int)prime[0];
			rtn /= p;
			rtn *= p-1;
		}
		return rtn;
	}
	
	//エラトステネスの篩
	//要素数がn+1のboolean配列を返す。
	//素数のindexはtrueとなり、それ以外はfalseとなる
	static boolean[] eratosthenes(int n){
		boolean[] flg = new boolean[n+1];
		Arrays.fill(flg, true);
		flg[0] = flg[1] = false;
		for(int i=2; i<n+1; i++){
			if(!flg[i])continue;
			for(int j=i*2; j<n+1; j += i)flg[j] = false;
		}
		return flg;
	}
	
	//素数判定
	static boolean isPrime(int n){
		int tmp = 2;
		boolean flg = true;
		while(tmp*tmp<=n){
			if (n%tmp == 0)flg = false;
			tmp++;
		}
		if(n <= 1)flg = false;
		return flg;
	}
	
	//拡張ユークリッドの互除法
	//返り値、aとbの最大公約数
	//X[0]とY[0]が、ax+by=gcd(a,b)を満たすようになる
	static long extGCD(long a, long b, long[] X, long[] Y){
		if(b == 0){
			X[0] = 1;
			Y[0] = 0;
			return a;
		}
		long d = extGCD(b, a%b, Y, X);
		Y[0] -= a/b * X[0];
		
		return d;
	}
	
	//２点間の距離
	public static double distance(double X1, double Y1, double X2, double Y2){
		return Math.sqrt((X1-X2)*(X1-X2) + (Y1-Y2)*(Y1-Y2));
	}
	
	//階乗
	public static int fact(int a){
		if(a == 0 || a == 1)return 1;
		return a*fact(a-1);
	}
		
	//約数の配列を返す
	static long[] divisor(long num){
		Set<Long> divisors = new TreeSet<>();
		long tmp=1;
		while(tmp*tmp<=num){
			if(num%tmp == 0){
				divisors.add(tmp);
				divisors.add(num/tmp);
			}
			tmp++;
		}
		Long[] rtn = new Long[divisors.size()];
		divisors.toArray(rtn);
		long[] ans = new long[rtn.length];
		for(int i=0; i<rtn.length; i++){
			ans[i] = rtn[i];
		}
		return ans;
	}
	
	//GCD最大公約数を返す
	static long gcd(long a, long b){
		if(b == 0) return a;
		return gcd(b, a%b);
	}

	static BigInteger biggcd(BigInteger a, BigInteger b){
		if(b == BigInteger.ZERO) return a;
		return biggcd(b, a.remainder(b));
	}
	
	//Bellman-Ford法
	//辺情報とstartの頂点を指定すると、すべての頂点への最短コストを返す（負の重みの辺もOK）
	//(問題に応じて要変化)1→Vのパスに負の閉路がある場合はindex:Vの要素に-1を入れて返す。
	static long[] bellmanFord(List<E> Edges, int from){
		long[] rtn = new long[V+1];
		Arrays.fill(rtn, INF);
		rtn[from] = 0;
		
		for(int i=0; i<V-1; i++){
			boolean update = false;
			for(E e : Edges){
				if(rtn[e.from] != INF && rtn[e.to] > rtn[e.from] + e.weight){
					rtn[e.to] = rtn[e.from] + e.weight;
					update = true;
				}
			}
			if(!update)break;
		}	
		//閉経路判定
		boolean[] is_update = new boolean[V];
		Arrays.fill(is_update, false);
		for(int i=0; i<V; i++){
			for(E e : Edges){
				if(rtn[e.from] != INF && rtn[e.to] > rtn[e.from] + e.weight){
					rtn[e.to] = rtn[e.from] + e.weight;
					is_update[e.to] = true;
				}
			}
		}
		
		if(is_update[V-1]){
			rtn[V] = -1;
		}
		return rtn;
	}
	
	//Kruskal法
	//最小全域木を構成したときのコストの総和を返す
	static long kruskal(List<E> Edges){
		Collections.sort(Edges);
		UnionFind uni = new UnionFind(V);
		long res = 0L;
		for (E e : Edges){
			if(!uni.same(e.from, e.to)){
				uni.join(e.from, e.to);
				res += e.weight;
			}
		}
		return res;
	}
	
	//Dijkstra法
	//辺情報とstartの頂点を指定すると、すべての頂点への最短コストを返す
	static long[] dijkstra(List<List<E>> G, int from){
		int N = G.size();
		long[] rtn = new long[N];
		Arrays.fill(rtn, INF);
		rtn[from] = 0;
		Queue<V> que = new PriorityQueue<>();
		
		que.add(new V(from,0));
        while(!que.isEmpty()){
            V doneNode = que.poll();
            for(E e: G.get(doneNode.num)){
                  long alt = rtn[doneNode.num] + e.weight;
                  if(rtn[e.to] > alt){
                        rtn[e.to] = alt;
                        que.add(new V(e.to, alt));
                  }     
            }
        }
		return rtn;
	}
	
	//グラフ用：辺
	static class E implements Comparable<E>{
		int from;
		int to;
		long weight;
		public E(int from, int to, long weight){
			this.from = from;
			this.to = to;
			this.weight = weight;
		}
		@Override
		public int compareTo(E o) {
			// TODO Auto-generated method stub
			return Long.compare(this.weight, o.weight);
		}
	}
	//グラフ用：頂点
	static class V implements Comparable<V>{
		int num;
		long cost;
		public V(int num, long cost){
			this.num = num;
			this.cost = cost;
		}
		//コストの低い順に並べる(PriorityQueue用)
		@Override
		public int compareTo(V o) {
			return Long.compare(this.cost, o.cost);
		}
	}
	
	//Union-Find木
	static class UnionFind{
		private int[] roots;
		public UnionFind(int N){
			this.roots = new int[N];
			for(int i=0; i<N; i++){
				this.roots[i] = i;
			}
		}
		
		public boolean same(int a, int b){
			if(root(a) == root(b)){
				return true;
			}else{
				return false;
			}
		}
		
		public int root(int a){
			if(a == this.roots[a])return a;
			return this.roots[a] = root(this.roots[a]);
		}
		
		public void join(int a, int b){
			if(root(a) == root(b))return;
			this.roots[root(b)] = root(a);
		}
		
		public int[] getRoots(){
			return this.roots;
		}
	}
 
	//素因数分解の結果AとBに対して、A*Bの素因数分解の結果を返す
	static List<long[]> factorizationMerge(List<long[]> A, List<long[]> B){
		List<long[]> primes = new ArrayList<>(A);
		for(long[] b : B){
			boolean isExist = false;
			for(long[] a : primes){
				if(a[0] == b[0]){
					isExist = true;
					a[1] += b[1]; 
				}
			}
			if(!isExist)primes.add(b);
		}
		return primes;
	}
	
	//素因数分解をする(index_0:素数 index_1:個数)
	static List<long[]> factorization(long M){
		List<long[]> primes = new ArrayList<>();
 
		long now_prime = 2;
		long now_M = M;
		while(now_prime*now_prime<=M){
			if(now_M%now_prime == 0){
				now_M /= now_prime;
				long[] prime = new long[2];
				prime[0] = now_prime;
				prime[1] = 1;
				while(now_M%now_prime == 0){
					now_M /= now_prime;
					prime[1]++;
				}
				primes.add(prime);
				
			}
			if(now_prime%2 == 0){
				now_prime++;
			}else{
				now_prime += 2;
			}
		}
		if (now_M>1){
			long[] prime = new long[2];
			prime[0] = now_M;
			prime[1] = 1;
			primes.add(prime);
		}
		
		return primes;
	}
	
	//２つのint配列を連結させる
	static int[] joinArray(int[] A, int[] B){
		int[] tmp = new int[A.length + B.length];
		System.arraycopy(A, 0, tmp, 0, A.length);
		System.arraycopy(B, 0, tmp, A.length, B.length);
		return tmp;
	}
	
	//２つのint配列が等しいかどうか判定する
	static boolean arrayEqual(int[] A, int[] B){
		if(A.length != B.length)return false;
		for(int i=0; i<A.length; i++){
			if(A[i] != B[i])return false;
		}
		return true;
	}
	
	//int配列をindex順に見て、要素の重複があった場合その要素と間の要素を削除する12123→23
	static int[] calc(int[] A){
		boolean[] tmp = new boolean[A.length];
		Arrays.fill(tmp, true);
		int fetch = 0;
		while(fetch<A.length-1){
			boolean isExist = false;
			for(int i=fetch+1; i<A.length;i++){
				if(A[fetch] == A[i]){
					for(int j=fetch;j<=i;j++){
						tmp[j]=false;
					}
					fetch = i+1;
					isExist = true;
					break;
				}
			}
			if(!isExist)fetch++;
		}
		List<Integer> list = new ArrayList<>();
		for(int i=0; i<A.length; i++){
			if(tmp[i])list.add(A[i]);
		}
		
		int[] foo = new int[list.size()];
		for(int i=0; i<list.size(); i++){
			foo[i] = list.get(i);
		}
		/*
		for(boolean t : tmp)
			System.out.print(t+" ");
		*/
		return foo;
	}
		
	//aとbの最小公倍数を返す
	static long lcm (long a, long b) {
		return ((a/gcd(a,b))*b);
	}
	
	//二項係数用の前処理
	static void COMinit(){
		//二項係数演算用前処理
		fac = new long[MAX+1];
		fac[0] = 1L;
		fac[1] = 1L;
		inv = new long[MAX+1];
		inv[1] = 1L;
		finv = new long[MAX+1];
		finv[0] = 1L;
		finv[1] = 1L;
		for(int i=2;i<MAX;i++){
			fac[i] = fac[i-1] * i % MOD;
			inv[i] = MOD - inv[(int)(MOD%i)] * (MOD/i) % MOD;
			finv[i] = finv[i-1] * inv[i] % MOD;
		}
		isPreprocessing = true;
	}
	
	//二項係数i_C_jをMODで割った余りを返す
	static long COM(int i, int j){
		if(i<j)return 0;
		if(i<0 || j<0)return 0;
		if(!isPreprocessing)COMinit();
		return fac[i] * (finv[j] * finv[i-j] % MOD) % MOD;
	}
	
	//入力
	//https://qiita.com/p_shiki37/items/a0f6aac33bf60f5f65e4
	static class FastScanner {
	    private final InputStream in = System.in;
	    private final byte[] buffer = new byte[1024];
	    private int ptr = 0;
	    private int buflen = 0;
	    private boolean hasNextByte() {
	        if (ptr < buflen) {
	            return true;
	        }else{
	            ptr = 0;
	            try {
	                buflen = in.read(buffer);
	            } catch (IOException e) {
	                e.printStackTrace();
	            }
	            if (buflen <= 0) {
	                return false;
	            }
	        }
	        return true;
	    }
	    private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
	    private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
	    public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
	    public String next() {
	        if (!hasNext()) throw new NoSuchElementException();
	        StringBuilder sb = new StringBuilder();
	        int b = readByte();
	        while(isPrintableChar(b)) {
	            sb.appendCodePoint(b);
	            b = readByte();
	        }
	        return sb.toString();
	    }
	    public long nextLong() {
	        if (!hasNext()) throw new NoSuchElementException();
	        long n = 0;
	        boolean minus = false;
	        int b = readByte();
	        if (b == '-') {
	            minus = true;
	            b = readByte();
	        }
	        if (b < '0' || '9' < b) {
	            throw new NumberFormatException();
	        }
	        while(true){
	            if ('0' <= b && b <= '9') {
	                n *= 10;
	                n += b - '0';
	            }else if(b == -1 || !isPrintableChar(b)){
	                return minus ? -n : n;
	            }else{
	                throw new NumberFormatException();
	            }
	            b = readByte();
	        }
	    }
	    public int nextInt() {
	        long nl = nextLong();
	        if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
	        return (int) nl;
	    }
	    public double nextDouble() { return Double.parseDouble(next());}
	}
}