#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

ll n;
ll arr[101010];
ll tree[202020];

void update(int x){
	while(x < 202020){
		tree[x]++;
		x += x & -x;
	}
}

int query(int x){
	int ret = 0;
	while(x){
		ret += tree[x];
		x -= x & -x;
	}
	return ret;
}

bool chk(int m){
	memset(tree, 0, sizeof tree);
	ll res = 0, t = n + 1;
	update(t);
	for(int i=1; i<=n; i++){
		t += arr[i] >= m ? 1 : -1;
		res += query(t);
		update(t);
	}
	return res >= ((n*n+n)/2+1)/2;
}

int main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n;
	for(int i=1; i<=n; i++) cin >> arr[i];
	
    int l = 0, r = 1e9;
	while(l < r){
		int m = l + r + 1 >> 1;
		if(chk(m)) l = m;
		else r = m - 1;
	}
	cout << l;
}
### Prompt

Please provide a CPP coded solution 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;

ll n;
ll arr[101010];
ll tree[202020];

void update(int x){
	while(x < 202020){
		tree[x]++;
		x += x & -x;
	}
}

int query(int x){
	int ret = 0;
	while(x){
		ret += tree[x];
		x -= x & -x;
	}
	return ret;
}

bool chk(int m){
	memset(tree, 0, sizeof tree);
	ll res = 0, t = n + 1;
	update(t);
	for(int i=1; i<=n; i++){
		t += arr[i] >= m ? 1 : -1;
		res += query(t);
		update(t);
	}
	return res >= ((n*n+n)/2+1)/2;
}

int main(){
    ios_base::sync_with_stdio(0); cin.tie(0);
    cin >> n;
	for(int i=1; i<=n; i++) cin >> arr[i];
	
    int l = 0, r = 1e9;
	while(l < r){
		int m = l + r + 1 >> 1;
		if(chk(m)) l = m;
		else r = m - 1;
	}
	cout << l;
}
```