#include <iostream>

using namespace std;
typedef long long ll;
ll n,N = 200020,bit[200020],a[100010];
ll sum(int i){
	ll s = 0;
	while(i>=0){
		s += bit[i];
		i = (i&(i+1)) -1;
	}
	return s;
}

ll add(int i, int x){
	while(i<=N){
		bit[i] += x;
		i |= i+1;
	}
}

bool check(ll x){
    for(int i=0;i<N;i++) bit[i] = 0;
    ll s = 100000,t = 0;
    add(s,1);
    for(int i=0;i<n;i++){
        if(a[i]>=x) s += 1;
        else s -= 1;
        t += sum(s);
        add(s,1);
    }
    if(t>=n*(n+1)/4) return true;
    else return false;
}

int main(){
    cin >> n;
    for(int i=0;i<n;i++){
        cin >> a[i];
    }
    if(n==1){
        cout << a[0] << endl;
        return 0;
    }
    ll l = 0,r = 1000000001;
    while(r-l>1){
        ll mid = (l+r)/2;
        if(check(mid)) l = mid;
        else r = mid;
    }
    cout << l << endl;
}
### Prompt

Your task is to create a cpp solution to the following problem:

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 <iostream>

using namespace std;
typedef long long ll;
ll n,N = 200020,bit[200020],a[100010];
ll sum(int i){
	ll s = 0;
	while(i>=0){
		s += bit[i];
		i = (i&(i+1)) -1;
	}
	return s;
}

ll add(int i, int x){
	while(i<=N){
		bit[i] += x;
		i |= i+1;
	}
}

bool check(ll x){
    for(int i=0;i<N;i++) bit[i] = 0;
    ll s = 100000,t = 0;
    add(s,1);
    for(int i=0;i<n;i++){
        if(a[i]>=x) s += 1;
        else s -= 1;
        t += sum(s);
        add(s,1);
    }
    if(t>=n*(n+1)/4) return true;
    else return false;
}

int main(){
    cin >> n;
    for(int i=0;i<n;i++){
        cin >> a[i];
    }
    if(n==1){
        cout << a[0] << endl;
        return 0;
    }
    ll l = 0,r = 1000000001;
    while(r-l>1){
        ll mid = (l+r)/2;
        if(check(mid)) l = mid;
        else r = mid;
    }
    cout << l << endl;
}
```