#include <bits/stdc++.h>
using namespace std;
using lld = int64_t;
const int N = 100000 + 5;

int a[ N ], s[ N ];
lld bit[ N << 1 ];

lld query( int p ) {
	lld ret = 0;
	while ( p ) {
		ret += bit[ p ];
		p -= p & -p;
	}
	return ret;
}

void add( int p ) {
	while ( p < N + N ) {
		bit[ p ]++;
		p += p & -p;
	}
}

lld count( int m, int n ) {
	for ( int i = 1 ; i <= n ; ++ i )
		if ( a[ i ] >= m ) s[ i ] = s[ i - 1 ] + 1;
		else s[ i ] = s[ i - 1 ] - 1;
	memset( bit, 0, sizeof( bit ) );
	lld ret = 0;
	for ( int i = 0 ; i <= n ; ++ i ) {
		ret += query( s[ i ] + N );
		add( s[ i ] + N );
	}
	return ret;
}

int main() {
	int n; scanf( "%d", &n );
	for ( int i = 1 ; i <= n ; ++ i )
		scanf( "%d", a + i );
	lld tot = ( ( lld )( n ) ) * ( n + 1 ) / 2;
	int l = 1, r = 1000000001;
	while ( r - l > 1 ) {
		int m = ( l + r ) >> 1;
		if ( count( m, n ) >= ( tot + 1 ) / 2 ) l = m;
		else r = m;
	}
	cout << l << '\n';
	return 0;
}
### Prompt

In CPP, your task is to solve 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;
using lld = int64_t;
const int N = 100000 + 5;

int a[ N ], s[ N ];
lld bit[ N << 1 ];

lld query( int p ) {
	lld ret = 0;
	while ( p ) {
		ret += bit[ p ];
		p -= p & -p;
	}
	return ret;
}

void add( int p ) {
	while ( p < N + N ) {
		bit[ p ]++;
		p += p & -p;
	}
}

lld count( int m, int n ) {
	for ( int i = 1 ; i <= n ; ++ i )
		if ( a[ i ] >= m ) s[ i ] = s[ i - 1 ] + 1;
		else s[ i ] = s[ i - 1 ] - 1;
	memset( bit, 0, sizeof( bit ) );
	lld ret = 0;
	for ( int i = 0 ; i <= n ; ++ i ) {
		ret += query( s[ i ] + N );
		add( s[ i ] + N );
	}
	return ret;
}

int main() {
	int n; scanf( "%d", &n );
	for ( int i = 1 ; i <= n ; ++ i )
		scanf( "%d", a + i );
	lld tot = ( ( lld )( n ) ) * ( n + 1 ) / 2;
	int l = 1, r = 1000000001;
	while ( r - l > 1 ) {
		int m = ( l + r ) >> 1;
		if ( count( m, n ) >= ( tot + 1 ) / 2 ) l = m;
		else r = m;
	}
	cout << l << '\n';
	return 0;
}
```