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: Wrong Answer)

#include<iostream>
#include<vector>
#include<string>
#include<array>
#include<cmath>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<numeric>
#include<iomanip>
#include<utility>
#include<cstdlib>
#include<typeinfo>
#include<stack>
#include<cassert>
#include<functional>
#define int long long int
#define double long double
using namespace std;
#define INT_MAX 1000000007
#define INF 1000000000000000007
const int MAX_N = 1 << 17;
#define REP(i,n) for(int (i)=0;(i)<(n);(i)++)
using namespace std;

signed main()
{
int n;
std::cin >> n;
std::vector<double> a(n,0);
double mean;
for(int i = 0; i < n; i++){
    double temp;
    std::cin >> temp;
    a[i] = temp;
    mean += temp;
}
mean = (double)mean/((double)n * 1.0);

double min = 1000000.00;
double flag = 0;
for(int i = 0; i < n; i++){
    if(abs(a[i] - mean) < min){
        min =abs(a[i] - mean);
        flag= i; 
    }
}

std::cout << flag << std::endl;
return 0;
}