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;

#define llong long long int
#define ldouble long double
#define rep(i, n) for (int i = 0; i < n; ++i)
#define repr(i, n) for (int i = n; i >= 0; --i)
#define stl_rep(itr, x) for (auto itr = x.begin(); itr != x.end(); ++itr)
#define all(x) x.begin(), x.end()
#define allr(x) x.rbegin(), x.rend()

const static int mod = 1000000000 + 7;
const static int inf = INT_MAX / 2;
const static llong INF = LLONG_MAX / 2;
const static double eps = 1e-10;
const static int dx[] = {1, 0, -1, 0};
const static int dy[] = {0, 1, 0, -1};

template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return 1;} return 0;}
template<class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return 1;} return 0;}

signed main (int argc, char *argv[]) {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, k;
    cin >> n >> k;
    vector<llong> A(n), Cum(n + 1);
    Cum[0] = 0;
    rep(i, n) {
        cin >> A[i];
        Cum[i + 1] = Cum[i] + A[i];
    }

    vector<llong> P;
    for (int i = 0; i < n; ++i) {
        for (int j = i; j < n + 1; ++j) {
            P.push_back(Cum[j] - Cum[i]);
        }
    }
    sort(allr(P));

    vector<llong> cur = P;
    for (int k = 64; k >= 0; --k) {
        vector<llong> tmp;
        for (int e = 0; e < cur.size(); ++e) {
            if (cur[e] & (1LL << k)) tmp.push_back(cur[e]);
        }

        if (tmp.size() >= k) cur = tmp;
    }

    llong res = cur[0];
    rep(i, k) {
        res &= cur[i];
    }

    cout << res << endl;

    return 0;
}