#include<iostream>
#include<cstdio>
#define mid ((l+r)>>1)
#define read(n) scanf("%lld",&n)
using namespace std;
typedef long long ll;
const int MAXN = 1e5 + 5;
ll a[MAXN] ,  binary[MAXN];
ll pre[MAXN];
ll b[MAXN];
ll ans;
ll n;
void merge(ll *p,int l,int r)
{
	if(l == r)	return ;
	merge(p,l,mid);
	merge(p,mid+1,r);
	int tot = l;
	int pos1 = l , pos2 = mid + 1;
	while(pos1 <= mid && pos2 <= r)
	{
		if(p[pos1] > p[pos2])
		{
			b[tot++] = p[pos2++];
			ans += mid - pos1 + 1;
		}	
		else
			b[tot++] = p[pos1++];
	}
	while(pos1 <= mid)	b[tot++] = p[pos1++];
	while(pos2 <= r)	b[tot++] = p[pos2++];
	for(int i = l ; i <= r ; i++)
		p[i] = b[i];
}
bool check(ll x)
{
	ans = 0;
	pre[0] = 0;
	for(int i = 1 ; i <= n ; i++)
		pre[i] = pre[i-1] + (x >= a[i] ? -1 : 1) , ans += pre[i] < 0 ? 1 : 0;
	merge(pre,1,n);
	return ans <= (long long)(n * (n + 1) / 4);
}
int main()
{
	read(n);
	for(int i = 1 ; i <= n ; i++)
		read(a[i]) , binary[i] = a[i];
	merge(binary,1,n);
	ans = 0;
	int l = 1 , r = n;
	while(l + 1 < r)
	{
		if(check(binary[mid]))
			l = mid;
		else
			r = mid;
	}
	printf("%d\n",binary[r]); 
}
### Prompt

Your task is to create 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<iostream>
#include<cstdio>
#define mid ((l+r)>>1)
#define read(n) scanf("%lld",&n)
using namespace std;
typedef long long ll;
const int MAXN = 1e5 + 5;
ll a[MAXN] ,  binary[MAXN];
ll pre[MAXN];
ll b[MAXN];
ll ans;
ll n;
void merge(ll *p,int l,int r)
{
	if(l == r)	return ;
	merge(p,l,mid);
	merge(p,mid+1,r);
	int tot = l;
	int pos1 = l , pos2 = mid + 1;
	while(pos1 <= mid && pos2 <= r)
	{
		if(p[pos1] > p[pos2])
		{
			b[tot++] = p[pos2++];
			ans += mid - pos1 + 1;
		}	
		else
			b[tot++] = p[pos1++];
	}
	while(pos1 <= mid)	b[tot++] = p[pos1++];
	while(pos2 <= r)	b[tot++] = p[pos2++];
	for(int i = l ; i <= r ; i++)
		p[i] = b[i];
}
bool check(ll x)
{
	ans = 0;
	pre[0] = 0;
	for(int i = 1 ; i <= n ; i++)
		pre[i] = pre[i-1] + (x >= a[i] ? -1 : 1) , ans += pre[i] < 0 ? 1 : 0;
	merge(pre,1,n);
	return ans <= (long long)(n * (n + 1) / 4);
}
int main()
{
	read(n);
	for(int i = 1 ; i <= n ; i++)
		read(a[i]) , binary[i] = a[i];
	merge(binary,1,n);
	ans = 0;
	int l = 1 , r = n;
	while(l + 1 < r)
	{
		if(check(binary[mid]))
			l = mid;
		else
			r = mid;
	}
	printf("%d\n",binary[r]); 
}
```