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 <iostream>
#include <map>
#include <algorithm>
#include <string>
#include <vector> 
#include <numeric>
#include <cmath>
#include <queue>
#include <iomanip>
#include <functional>

#define CHMAX(a,b) a = std::max(a,b)
#define CHMIN(a,b) a = std::min(a,b)
#define MAXMIN(a,b,c) std::max(a,min(b,c))
#define CHABS(a) a = std::abs(a) 
#define COUT(a) std::cout << a << std::endl
#define CERR(a) std::cerr << a << std::endl
#define FOR(n) for(lli i = 0; i < n; i++)

using namespace std;
using lli = long long int;
using pll = pair<lli, lli>;
using tlll = tuple<lli, lli, lli>;
using vll = vector<lli>;

lli mod197 = 1000000007LL;
lli INF = 10000000000000;

// ax + by = gcd(a,b) 最大公約数
template< typename T >
T extgcd(T a, T b, T& x, T& y) {
	T d = a;
	if (b != 0) {
		d = extgcd(b, a % b, y, x);
		y -= (a / b) * x;
	}
	else {
		x = 1;
		y = 0;
	}
	return d;
}

int main(void) {

	int n;
	cin >> n;

	double ave = 0;
	int arr[100];
	for (int i = 0; i < n; i++) {
		cin >> arr[i];
		ave += arr[i];
	}

	ave /= n;
	int ans_num = 0;
	double mini = 1000;

	for (int i = 0; i < n; i++) {
		if (mini > abs(ave - arr[i])) {
			mini = abs(ave - arr[i]);
			ans_num = i;
		}
	}

	COUT(ans_num);

	return 0;
}