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>
#define rep(i, a, n) for(int (i) = (int)(a);(i) < (int)(n);++(i))
#define repe(i, a, n) for(int (i) = (int)(a);(i) <= (int)(n);++(i))
#define repif(i, a, x) for(int (i) = (int)(a);(x);++(i))
#define repr(i, a, n) for(int (i) = ((int)(a)-1);(i) >= (int)(n);--(i))
#define reper(i, a, n) for(int (i) = (int)(a);(i) >= (int)(n);--(i))
#define chmax(a, b) a = max(a, b)
#define chmin(a, b) a = min(a, b)
#define SIZE(x) ((int)(x).size())
#define ALL(x) (x).begin(), (x).end()
#define RNG(x, a, n) &((x)[a]), &((x)[n])
#define ADD(a, b) ((((a+MOD)%MOD)+((b+MOD)%MOD))%MOD)
#define MUL(a, b) ((((a)%MOD)*((b)%MOD))%MOD)
#define CEIL(x) ((int)ceil(((double)x)))
#define POW(x, y) ((int)pow(x, y))
#define UNIQUE(x) (x).erase(unique(ALL((x))), (x).end())
#define YES cout << "YES" << endl
#define NO cout << "NO" << endl
#define Yes cout << "Yes" << endl
#define No cout << "No" << endl
#define yes cout << "yes" << endl;
#define no cout << "no" << endl;
#define fcout cout << fixed << setprecision(10)
#define pb push_back
#define pf push_front
#define mp make_pair
#define mt make_tuple
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define PI 3.1415926535897
#define EPS (1e-10)
#define INF INT_MAX
#define INFLL LLONG_MAX
#define MOD 1000000007
using llint = long long;
using VI = std::vector<int>;
using PII = std::pair<int, int>;
using SI = std::set<int>;
using MII = std::map<int, int>;
using TIII = std::tuple<int, int, int>;
using MCI = std::map<char, int>;
using VPII = std::vector<PII>;
using namespace std;


int N;
int a[110];


signed main(){
	cin >> N;
	rep(i, 0, N) cin >> a[i];

	double av = accumulate(RNG(a, 0, N), 0)/(double)N;

	double err = INF;
	int ans = 0;

	rep(i, 0, N)
		if(abs(a[i]-av) < err) ans = i, err = abs(a[i]-av);

	cout << ans << endl;

	return 0;
}