Problem:
Score : 200 points

Problem Statement

Niwango-kun is an employee of Dwango Co., Ltd.

One day, he is asked to generate a thumbnail from a video a user submitted.

To generate a thumbnail, he needs to select a frame of the video according to the following procedure:

Get an integer N and N integers a_0, a_1, ..., a_{N-1} as inputs. N denotes the number of the frames of the video, and each a_i denotes the representation of the i-th frame of the video.

Select t-th frame whose representation a_t is nearest to the average of all frame representations.

If there are multiple such frames, select the frame with the smallest index.

Find the index t of the frame he should select to generate a thumbnail.

Constraints

1 \leq N \leq 100

1 \leq a_i \leq 100

All numbers given in input are integers

Input

Input is given from Standard Input in the following format:

N
a_{0} a_{1} ... a_{N-1}

Output

Print the answer.

Sample Input 1

3
1 2 3

Sample Output 1

1

Since the average of frame representations is 2, Niwango-kun needs to select the index 1, whose representation is 2, that is, the nearest value to the average.

Sample Input 2

4
2 5 2 5

Sample Output 2

0

The average of frame representations is 3.5.

In this case, every frame has the same distance from its representation to the average.

Therefore, Niwango-kun should select index 0, the smallest index among them.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

//#include <bits/stdc++.h>
//#include <stdio.h>
#include<iostream>
#include<cstdio>
#include<bitset>
#include<algorithm>
#include<vector>
#include<list>
#include<queue>
#include<stack>
#include<string>
#include<string.h>
#include<cmath>
#include<utility>
#include<functional>
#include<map>
#include<set>
#include<cctype>
#include<fstream>


#define FOR(i, a, b) for( int i=(a);i<=(b);i++)
#define RFOR(i, a, b) for( int i=(a);i>=(b);i--)
#define LFOR(i, a, b) for( long long int i=(a);i<=(b);i++)
#define LRFOR(i, a, b) for(long long int i=(a);i>=(b);i--)
#define MOD 1000000007
#define INF 1000000000 //2000000000
#define LLINF 1000000000000000000 //9000000000000000000
#define PI 3.14159265358979
#define MAXI 7500000

using namespace std;
typedef long long int ll;
typedef pair< long long int, long long  int> P;

int dy[5] = { 0,0,1,-1,0 };
int dx[5] = { 1,-1,0,0 ,0 };


int main(void) {
	int n;
	int a[101];
	int pos = 0;
	double mini = INF;
	double ave = 0;

	cin >> n;
	FOR(i, 0, n-1) {
		cin >> a[i];
		ave += a[i];
	}
	ave /= (double)n;

	mini = fabs((double)a[0] - ave);

	FOR(i, 1, n - 1) {
		if (fabs((double)a[i] - ave) < mini) {
			mini = fabs((double)a[i] - ave);
			pos = i;
		}
	}
	cout << pos << endl;

	return 0;
}