We will define the median of a sequence b of length M, as follows:

* Let b' be the sequence obtained by sorting b in non-decreasing order. Then, the value of the (M / 2 + 1)-th element of b' is the median of b. Here, / is integer division, rounding down.



For example, the median of (10, 30, 20) is 20; the median of (10, 30, 20, 40) is 30; the median of (10, 10, 10, 20, 30) is 10.

Snuke comes up with the following problem.

You are given a sequence a of length N. For each pair (l, r) (1 \leq l \leq r \leq N), let m_{l, r} be the median of the contiguous subsequence (a_l, a_{l + 1}, ..., a_r) of a. We will list m_{l, r} for all pairs (l, r) to create a new sequence m. Find the median of m.

Constraints

* 1 \leq N \leq 10^5
* a_i is an integer.
* 1 \leq a_i \leq 10^9

Input

Input is given from Standard Input in the following format:


N
a_1 a_2 ... a_N


Output

Print the median of m.

Examples

Input

3
10 30 20


Output

30


Input

1
10


Output

10


Input

10
5 9 5 9 8 9 3 5 4 3


Output

8
#include <iostream>
#include <algorithm>
#include <string.h>

using namespace std;

typedef long long ll;
const int MAX = 1e5+1;

int N;

int a[MAX];
ll fenwick[2*MAX];
 
void update(ll x, ll v) {
  x++;
  for(; x <= 2 * N; x += x & (-x))
    fenwick[x] += v;
}
 
ll sum(ll x) {
  ll res = 0;
  x++;
  for(; x; x -= x & (-x)) 
    res += fenwick[x];
  return res;
}

bool ok(int x) {
  memset(fenwick, 0, sizeof(fenwick));
  ll ans = 0, total = 0ll;
  for (int i = 0; i < N; i++) {
    total += (a[i] >= x) ? 1 : -1;
    ans += sum(total+N) + ((total >= 0) ? 1 : 0);
    update(total+N, 1);
  }
  return ans >= ((1ll*N*(N+1))/2)/2 ;
}

int main () {
  cin >> N;
  for (int i = 0; i < N; i++) cin >> a[i];

  if (N == 1) {
    cout << a[0] << endl;
    return 0;
  }
  int l = 1, r = 1e9+1;
  while (r-l > 1) {
    int m = (l+r)/2;
    if (ok(m)) l = m;
    else r = m;
  }

  cout << l << endl;
  return 0;
}