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 <bits/stdc++.h>
#define rep(i, n) for(int i=0; i<(n); i++)
using namespace std;

int n;
int a[100010];
long bit[100010];

long sum(int i){
  long sum = 0;
  while(i > 0){
    sum += bit[i];
    i -= i & -i;
  }
  return sum;
}

void add(int i, long x){
  while(i <= n+1){
    bit[i] += x;
    i += i & -i;
  }
}

long iinv(int s[]){
  vector<int> t(n+1);
  rep(i, n+1) t[i] = s[i];
  sort(t.begin(), t.end());
  t.erase(unique(t.begin(), t.end()), t.end());
  map<int, int> m;
  rep(i, t.size()) m[t[i]] = i+1;
  rep(i, n+1) s[i] = m[s[i]];
  rep(i, n+2) bit[i] = 0;
  long ans = 0;
  rep(i, n+1){
    ans += sum(s[i]);
    add(s[i], 1);
  }
  return ans;
}   

bool check(int mid){
  int s[n+1];
  s[0] = 0;
  rep(i, n){
    int t;
    if(a[i] < mid) t = -1;
    else t = 1;
    s[i+1] = s[i] + t;
  }
  long k = n;
  return iinv(s) >= ((k+1)*k/2 - 1) / 2 + 1;
}

int main(){
  cin >> n;
  rep(i, n) cin >> a[i];
  int ok = 0;
  int ng = 1000000001;
  while(ng-ok > 1){
    int mid = (ok+ng) / 2;
    if(check(mid)) ok = mid;
    else ng = mid;
  }
  cout << ok << endl;
}