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


template <typename T>
struct BIT{
    int n;
    vector<T> d;

    BIT(int sz){
        n = sz+10;
        d.resize(n);
    }

    void add(int i, T x){

        while(i<=n){
            d[i] += x;
            i += i&-i;
        }
    }

    T getSum(int i){
        T ret = 0;

        while(i>0){
            ret += d[i];
            i -= i&-i;
        }
        return ret;
    }

    T getSum(int a, int b){
        return getSum(b-1) - getSum(a-1);
    }
};


int main(){
    ll N; cin >> N;
    vector<ll> a(N);
    rep(i,N) cin >> a[i];

    ll ng = 0, ok = 1e9;
    while(ok-ng>1){
        ll mid = (ng+ok)/2;

        ll sum = 0, num = 0;
        BIT<ll> bit(N*2+10);
        bit.add(N+1,1);
        rep(i,N){
            if(a[i]<=mid) sum++;
            else sum--;
            num += bit.getSum(1,sum+N+1);
            bit.add(sum+N+1,1);
        }
        if (num > (N+1)*N/2/2) ok = mid;
        else ng = mid;
    }

    cout << ok << endl;
}
### Prompt

Develop a solution in cpp to the problem described below:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i,n) for(ll i=0;i<(ll)n;i++)


template <typename T>
struct BIT{
    int n;
    vector<T> d;

    BIT(int sz){
        n = sz+10;
        d.resize(n);
    }

    void add(int i, T x){

        while(i<=n){
            d[i] += x;
            i += i&-i;
        }
    }

    T getSum(int i){
        T ret = 0;

        while(i>0){
            ret += d[i];
            i -= i&-i;
        }
        return ret;
    }

    T getSum(int a, int b){
        return getSum(b-1) - getSum(a-1);
    }
};


int main(){
    ll N; cin >> N;
    vector<ll> a(N);
    rep(i,N) cin >> a[i];

    ll ng = 0, ok = 1e9;
    while(ok-ng>1){
        ll mid = (ng+ok)/2;

        ll sum = 0, num = 0;
        BIT<ll> bit(N*2+10);
        bit.add(N+1,1);
        rep(i,N){
            if(a[i]<=mid) sum++;
            else sum--;
            num += bit.getSum(1,sum+N+1);
            bit.add(sum+N+1,1);
        }
        if (num > (N+1)*N/2/2) ok = mid;
        else ng = mid;
    }

    cout << ok << endl;
}
```