G: Working

Kou decided to do the same number of jobs every day for the next $ N $.

$ A_i $ jobs are added on the $ i $ day of the $ N $ day.

Mr. Kou has no work to do now, and he doesn't have to finish all the work by the $ N $ day.

How many jobs can you do in a day?

However, Mr. Kou is excellent, so he can do as many jobs as he has.

input

$ N $ is given on the first line.

On the second line, $ N $ integers $ A_1, A_2, A_3, \ dots, A_N $ are given, separated by blanks.

output

Output the maximum number of jobs you can do in a day. Insert a line break at the end.

Constraint

* $ N $ is an integer greater than or equal to $ 1 $ and less than or equal to $ 100 $
* $ A_1, A_2, A_3, \ dots, A_N $ are integers between $ 1 $ and $ 100 $



Input example 1


Five
4 2 5 3 1


Output example 1


3


If you decide to work more than $ 4 $ a day, you'll run out of work on the second day.

Input example 2


Five
9 9 1 9 9


Output example 2


6






Example

Input

5
4 2 5 3 1


Output

3
#include <bits/stdc++.h>
using namespace std;

const int INF = 1e9;
const int MOD = 1e9 + 7;

#define int				long long
#define FOR( i, m, n )  for( int (i) = (m); (i) < (n); (i)++ )
#define FORE( i, m, n ) FOR( i, m , n + 1 )
#define REP( i, n )     FOR( i, 0, n )
#define REPE( i, n )	FOR( i, 0, n + 1 )
#define FORR( i, m, n ) for( int (i) = (m); (i) >= 0; (i)-- )
#define ALL( a )        (a).begin(), (a).end()             
#define MP( a, b )		make_pair( (a), (b) )
#define fi				first
#define se				second

signed main() {
	int n; cin >> n;
	vector<int> a( n );

	REP( i, n ) cin >> a[i];

	int sum = 0, ans = INF;
	REP( i, n ) {
		sum += a[i];
		ans = min( sum / ( i + 1 ), ans );
	}

	cout << ans << endl;
}
