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

int n ;

int a[100010] ;

int b[100010] ;

int tree[200010] ;

void update(int pos){
	while(pos <= 2 * n){
		tree[pos] ++ ;
		pos += pos & (-pos) ;
	}
}

int get(int pos){
	int ans = 0 ;
	while(pos){
		ans += tree[pos] ;
		pos -= pos & (-pos) ;
	}
	return ans; 
}

bool check(int value){
	long long total = 0 ;
	int cur = n ;
	update(n) ;
	for (int i = 1; i <= n; ++ i){
		if(a[i] <= value) cur ++ ;
		else cur -- ;
		total += 1ll * get(cur - 1) ;
		update(cur) ;
	}	
	memset(tree, 0, sizeof(tree)) ;
	long long limit = 1ll * n * (n + 1) / 2 ;
	limit = (limit + 2ll) / 2 ;
	return total >= limit ;
}

int main(){
	#define NAME "MEDIAN"
	//freopen( NAME".inp" , "r" , stdin ) ; freopen( NAME".out" , "w" , stdout ) ;

	ios_base :: sync_with_stdio(false) ;

	cin >> n ;

	for (int i = 1; i <= n; ++ i) cin >> a[i], b[i] = a[i] ;

	sort(b + 1, b + n + 1) ;
	
	int l = 1, r = n ;

	while(l < r){
		int mid = (l + r) / 2 ;
		if(check(b[mid])) r = mid ;
		else l = mid + 1 ;
	}

	cout << b[l] << endl ;

	return 0 ;
}
### Prompt

Develop a solution in Cpp 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 ;

int n ;

int a[100010] ;

int b[100010] ;

int tree[200010] ;

void update(int pos){
	while(pos <= 2 * n){
		tree[pos] ++ ;
		pos += pos & (-pos) ;
	}
}

int get(int pos){
	int ans = 0 ;
	while(pos){
		ans += tree[pos] ;
		pos -= pos & (-pos) ;
	}
	return ans; 
}

bool check(int value){
	long long total = 0 ;
	int cur = n ;
	update(n) ;
	for (int i = 1; i <= n; ++ i){
		if(a[i] <= value) cur ++ ;
		else cur -- ;
		total += 1ll * get(cur - 1) ;
		update(cur) ;
	}	
	memset(tree, 0, sizeof(tree)) ;
	long long limit = 1ll * n * (n + 1) / 2 ;
	limit = (limit + 2ll) / 2 ;
	return total >= limit ;
}

int main(){
	#define NAME "MEDIAN"
	//freopen( NAME".inp" , "r" , stdin ) ; freopen( NAME".out" , "w" , stdout ) ;

	ios_base :: sync_with_stdio(false) ;

	cin >> n ;

	for (int i = 1; i <= n; ++ i) cin >> a[i], b[i] = a[i] ;

	sort(b + 1, b + n + 1) ;
	
	int l = 1, r = n ;

	while(l < r){
		int mid = (l + r) / 2 ;
		if(check(b[mid])) r = mid ;
		else l = mid + 1 ;
	}

	cout << b[l] << endl ;

	return 0 ;
}
```