Problem:
Score: 400 points

Problem Statement

To become a millionaire, M-kun has decided to make money by trading in the next N days. Currently, he has 1000 yen and no stocks - only one kind of stock is issued in the country where he lives.

He is famous across the country for his ability to foresee the future. He already knows that the price of one stock in the next N days will be as follows:

A_1 yen on the 1-st day, A_2 yen on the 2-nd day, ..., A_N yen on the N-th day.

In the i-th day, M-kun can make the following trade any number of times (possibly zero), within the amount of money and stocks that he has at the time.

Buy stock: Pay A_i yen and receive one stock.

Sell stock: Sell one stock for A_i yen.

What is the maximum possible amount of money that M-kun can have in the end by trading optimally?

Constraints

2 \leq N \leq 80

100 \leq A_i \leq 200

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 \cdots A_N

Output

Print the maximum possible amount of money that M-kun can have in the end, as an integer.

Sample Input 1

7
100 130 130 130 115 115 150

Sample Output 1

1685

In this sample input, M-kun has seven days of trading. One way to have 1685 yen in the end is as follows:

Initially, he has 1000 yen and no stocks.

Day 1: Buy 10 stocks for 1000 yen. Now he has 0 yen.

Day 2: Sell 7 stocks for 910 yen. Now he has 910 yen.

Day 3: Sell 3 stocks for 390 yen. Now he has 1300 yen.

Day 4: Do nothing.

Day 5: Buy 1 stock for 115 yen. Now he has 1185 yen.

Day 6: Buy 10 stocks for 1150 yen. Now he has 35 yen.

Day 7: Sell 11 stocks for 1650 yen. Now he has 1685 yen.

There is no way to have 1686 yen or more in the end, so the answer is 1685.

Sample Input 2

6
200 180 160 140 120 100

Sample Output 2

1000

In this sample input, it is optimal to do nothing throughout the six days, after which we will have 1000 yen.

Sample Input 3

2
157 193

Sample Output 3

1216

In this sample input, it is optimal to buy 6 stocks in Day 1 and sell them in Day 2, after which we will have 1216 yen.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

//--------------------------------------------------------------------------------
// Code By:
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//					#####      ##       #####      ##       #   ###
//					##   #              #   #               #   # #
//					##    #	            #####               ##### #
//---------------------------------------------------------------------------------

#include<bits/stdc++.h>
using namespace std;

                             /* ----------------------------- START of Template ---------------*/
#define ll long long int
#define FOR(x,to) for(x=0;x<(to);x++)
#define ALL(a) (a.begin()),(a.end())
#define ZERO(a) memset(a,0,sizeof(a))
#define rep(i,a,b) for(ll i=a;i<b;++i)
#define repd(i,a,b) for(ll i=a;i>=b;--i)
#define mp make_pair
#define hell 1000000007
#define vvll vector<vector<long long int> >
#define vll vector<long long int>
#define MINUS(a) memset(a,0xff,sizeof(a))
#define SZ(X) ((int)(X).size())
#define REP(I, N) for (int I = 0; I < (N); ++I)
#define REPP(I, A, B) for (int I = (A); I < (B); ++I)
#define FORS(I, S) for (int I = 0; S[I]; ++I)
#define RS(X) scanf("%s", (X))
#define MP make_pair
#define PB push_back
#define MS0(X) memset((X), 0, sizeof((X)))
#define MS1(X) memset((X), -1, sizeof((X)))
#define LEN(X) strlen(X)
#define F first
#define S second

void getZarr(string str, vll &Z)
{
    ll n = str.length();
    ll L, R, k;
    L = R = 0;
    for (ll i = 1; i < n; ++i)
    {
        if (i > R)
        {
            L = R = i;
            while (R<n && str[R-L] == str[R])
                R++;
            Z[i] = R-L;
            R--;
        }
        else
        {
            k = i-L;
            if (Z[k] < R-i+1)
                Z[i] = Z[k];
            else
            {
                L = i;
                while (R<n && str[R-L] == str[R])
                    R++;
                Z[i] = R-L;
                R--;
   } } }}
vector<ll>isprime;
void seive(ll size){
	isprime.resize(size+1,1);
	ll i,j;
	isprime[0]=0;
	isprime[1]=0;
	for(i=2;i*i<=size;i++){
		if(isprime[i]){
			for(j=2;j*i<=size;j++)isprime[i*j]=0;
		}
	}
}
ll power(ll x,unsigned ll y, ll m){
	if(y==0)return 1;
	ll p=power(x,y/2,m);
	p=(p*p)%m;
	if(y%2!=0)p=(p*x)%m;
	return p;
}
ll modinverse(ll a,ll m){
	return power(a,m-2,m);
}


                       /* ----------------------          End Of Template --------------------------------*/

ll findans(){
	ll n,i,j;
	cin>>n;
	vll v(n);
	for(i=0;i<n;i++)cin>>v[i];
	vll vt;
	for(i=0;i<n;i++){
		if(vt.size()!=0){
			if(vt[vt.size()-1]==v[i])continue;
		}
		vt.PB(v[i]);
		
	}
	vll mini,maxi;
//	cout<<vt.size()<<" x\n";
	if(1<vt.size()){
		if(vt[0]<vt[1])mini.PB(vt[0]);
	}
	for(i=1;i<vt.size()-1;i++){
		if(vt[i]>vt[i-1]&&vt[i]>vt[i+1])maxi.PB(vt[i]);
		else if(vt[i]<vt[i-1]&&vt[i]<vt[i+1]) mini.PB(vt[i]);
	}
	ll ss=vt.size()-1;
	if(vt.size()>1){
		if(vt[ss]<vt[ss-1])mini.PB(vt[ss]);
		else maxi.PB(vt[ss]);
	}
	ll curr=1000;
	for(i=0;i<mini.size();i++){
		ll cnt=curr/mini[i];
		if(i>=maxi.size())break;
		curr-=cnt*mini[i];
		
		curr+=cnt*maxi[i];
	}
	cout<<curr<<endl;
	return 0;
	
	



}

int main(){
	ios_base :: sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
	ll t,i;
//	cin>>t;
t=1;
	for(i=0;i<t;i++){
		findans();
	}
return 0;
}