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

#define N 100000

long long n;
int a[N];

int a_sort[N];
int b[N];
int s[N];

class BIT {
	int m;
	vector<int> bit;

public:
	BIT( int n )
	{
		m = n;
		bit.resize(m + 1);
		fill( bit.begin(), bit.end(), 0 );
	}

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

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

int main()
{
	cin >> n;
	for( int i = 0; i < n; i++ ) {
		cin >> a[i];
		a_sort[i] = a[i];
	}

	sort( a_sort, a_sort + n );
	int lb = 0, ub = n;
	while( ub - lb > 1 ) {
		int x = ( lb + ub ) / 2;
		for( int i = 0; i < n; i++ ) {
			b[i] = ( a[i] < a_sort[x] ) ? -1 : 1;
		}
		for( int i = 0; i < n; i++ ) {
			s[i] = ( i == 0 ) ? b[i] : s[i - 1] + b[i];
		}
		long long cnt = 0;
		BIT bit( 2 * n + 1 );
		bit.add( n, 1 );
		for(int i = 0; i < n; i++ ) {
			cnt += bit.sum( s[i] + n );
			bit.add( s[i] + n, 1 );
		}
		if( cnt * 2 * 2 >= n * (n + 1) ) {
			lb = x;
		}
		else {
			ub = x;
		}
	}
	int ans = a_sort[ lb ];

	cout << ans << endl;

	return 0;
}

### Prompt

Create a solution in cpp for 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;

#define N 100000

long long n;
int a[N];

int a_sort[N];
int b[N];
int s[N];

class BIT {
	int m;
	vector<int> bit;

public:
	BIT( int n )
	{
		m = n;
		bit.resize(m + 1);
		fill( bit.begin(), bit.end(), 0 );
	}

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

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

int main()
{
	cin >> n;
	for( int i = 0; i < n; i++ ) {
		cin >> a[i];
		a_sort[i] = a[i];
	}

	sort( a_sort, a_sort + n );
	int lb = 0, ub = n;
	while( ub - lb > 1 ) {
		int x = ( lb + ub ) / 2;
		for( int i = 0; i < n; i++ ) {
			b[i] = ( a[i] < a_sort[x] ) ? -1 : 1;
		}
		for( int i = 0; i < n; i++ ) {
			s[i] = ( i == 0 ) ? b[i] : s[i - 1] + b[i];
		}
		long long cnt = 0;
		BIT bit( 2 * n + 1 );
		bit.add( n, 1 );
		for(int i = 0; i < n; i++ ) {
			cnt += bit.sum( s[i] + n );
			bit.add( s[i] + n, 1 );
		}
		if( cnt * 2 * 2 >= n * (n + 1) ) {
			lb = x;
		}
		else {
			ub = x;
		}
	}
	int ans = a_sort[ lb ];

	cout << ans << endl;

	return 0;
}

```