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<bits/stdc++.h>
using namespace std;
typedef long long unsigned int ll;

// definition {{{ 1

// scaning {{{ 2
#define Scd(x) scanf("%d", &x)
#define Scd2(x,y) scanf("%d%d", &x, &y)
#define Scd3(x,y,z) scanf("%d%d%d", &x, &y, &z)

#define Scll(x) scanf("%llu", &x)
#define Scll2(x,y) scanf("%llu%llu", &x, &y)
#define Scll3(x,y,z) scanf("%llu%llu%llu", &x, &y, &z)

#define Scc(c) scanf("%c", &c);
#define Scs(s) scanf("%s", s);
#define Scstr(s) scanf("%s", &s);
// }}} 2

// constants {{{ 2
#define EPS (1e-7)
#define INF (1e9)
#define PI (acos(-1))
// }}} 2

// systems {{{ 2
#define Rep(x,y) for(int x = 0; x < y; x++)
#define Repe(x,y,z) for(int x = z; x < y; x++)
// }}} 2

// output {{{ 2
#define YesNo(a) (a)?printf("Yes\n"):printf("No\n");
// }}} 2

// }}} 1

int main() {

    int n;
    int a[300];
    float sum = 0;

    Scd(n);
    Rep(i,n){
        Scd(a[i]);
        sum += a[i];
    }

    sum /= n;

    int tmp = INF;
    int ans = 0;

    for( int i = n-1; i >= 0; i-- ){
        if( tmp >= abs<float>( sum - (float)a[i] ) ){
            ans = i;
            tmp = abs<float>( sum - (float)a[i] );
        }
    }


    printf ("%d\n", ans);

    return 0;
}