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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
double ABS(double n) {
    return n<0?(-1)*n:n;
}

void sort(double *a, int size) {
    for(int i=0;i<size;i++) {
        for(int j=i;j<size-1;j++) {
            if (a[j] > a[j+1]) {
                double tmp = a[j];
                a[j] = a[j+1];
                a[j+1] = tmp;
            }
        }
    }
}

int main(void) {
    int N;
    scanf("%d", &N);
    int a[N];
    double b[N];
    double sum=0;
    for (int i=0;i<N;i++) {
        scanf("%d", &a[i]);
        b[i] = a[i];
        sum += b[i];
    }

    double mean = sum/N;
    for (int i=0;i<N;i++) {
        b[i] = ABS(b[i]-mean);
    }
    sort(b, N);
    int i=0;
    int ans=0;
    while(i<N) {
        if(ABS(a[i]-mean)==b[0]) {
            ans=i;
            break;
        }
        i++;
    }

    printf("%d\n", ans);
    return 0;
}