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

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

const int N = 100005;
const int M = 200005;
int n, a[N], f[M];

void update(int i, int x, int sz){
	// cout<<i<<endl;
	while(i <= sz){
		f[i] += x;
		i += i&(-i);
	}
}

int query(int i){
	int ret = 0;
	while(i){
		ret += f[i];
		i &= (i - 1);
	}
	return ret;
}

ll count(int m){
	int sz = 2*(n + 1);
	for(int i = 0; i <= sz; i++){
		f[i] = 0;
	}
	ll ret = 0;
	update(n + 1, 1, sz);
	int cur = 0;
	for(int i = 1; i <= n; i++){
		if(a[i] <= m)cur--;
		else cur++;
		ret += i - query(n + 1 + cur);
		update(n + 1 + cur, 1, sz);
 	}
 	return ret;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	cout<<setprecision(32);

	cin>>n;
	for(int i = 1; i <= n; i++){
		cin>>a[i];
	}
	ll req = n*1LL*(n + 1)/2;
	req = req/2 + 1;
	int lo = 1, hi = 1e9;
	while(lo < hi){
		int mi = lo + (hi - lo)/2;
		if(req <= count(mi))hi = mi;
		else lo = mi + 1;
	}
	cout<<lo<<'\n';


	return 0;
}
### Prompt

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

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;

const int N = 100005;
const int M = 200005;
int n, a[N], f[M];

void update(int i, int x, int sz){
	// cout<<i<<endl;
	while(i <= sz){
		f[i] += x;
		i += i&(-i);
	}
}

int query(int i){
	int ret = 0;
	while(i){
		ret += f[i];
		i &= (i - 1);
	}
	return ret;
}

ll count(int m){
	int sz = 2*(n + 1);
	for(int i = 0; i <= sz; i++){
		f[i] = 0;
	}
	ll ret = 0;
	update(n + 1, 1, sz);
	int cur = 0;
	for(int i = 1; i <= n; i++){
		if(a[i] <= m)cur--;
		else cur++;
		ret += i - query(n + 1 + cur);
		update(n + 1 + cur, 1, sz);
 	}
 	return ret;
}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
	cout<<setprecision(32);

	cin>>n;
	for(int i = 1; i <= n; i++){
		cin>>a[i];
	}
	ll req = n*1LL*(n + 1)/2;
	req = req/2 + 1;
	int lo = 1, hi = 1e9;
	while(lo < hi){
		int mi = lo + (hi - lo)/2;
		if(req <= count(mi))hi = mi;
		else lo = mi + 1;
	}
	cout<<lo<<'\n';


	return 0;
}
```