#include<map>
#include<set>
#include<stack>
#include<cmath>
#include<queue>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;

int a[100005], sum[100005], b[100005], n;
long long t1, t2;

void ms(int l, int r)
{
	if (l == r) return;
	int m = (l + r) >> 1;
	ms(l, m);
	ms(m + 1, r);
	int x = l, y = m + 1, z = l;
	while ((x <= m) && (y <= r))
	{
		if (sum[x] < sum[y]) b[z++] = sum[x++];
		else
		{
			t2 += x - l;
			b[z++] = sum[y++];
		}
	}
	while (x <= m) b[z++] = sum[x++];
	while (y <= r)
	{
		t2 += x - l;
		b[z++] = sum[y++];
	}
	for (int i = l; i <= r; i++) sum[i] = b[i];
}

bool check(int x)
{
	t2 = 0;
	for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + (a[i] <= x ? 1 : -1);
	ms(0, n);
	return t2 >= (t1 / 2 + 1);
}

int main()
{
	int l = inf, r = -inf;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
	{
		scanf("%d", &a[i]);
		l = min(l, a[i]);
		r = max(r, a[i]);
		t1 += i;
	}
	while (l < r)
	{
		int m = (l + r) >> 1;
		if (check(m)) r = m;
		else l = m + 1;
	}
	printf("%d", r);
	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<map>
#include<set>
#include<stack>
#include<cmath>
#include<queue>
#include<cstdio>
#include<string>
#include<vector>
#include<cstring>
#include<iostream>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;

int a[100005], sum[100005], b[100005], n;
long long t1, t2;

void ms(int l, int r)
{
	if (l == r) return;
	int m = (l + r) >> 1;
	ms(l, m);
	ms(m + 1, r);
	int x = l, y = m + 1, z = l;
	while ((x <= m) && (y <= r))
	{
		if (sum[x] < sum[y]) b[z++] = sum[x++];
		else
		{
			t2 += x - l;
			b[z++] = sum[y++];
		}
	}
	while (x <= m) b[z++] = sum[x++];
	while (y <= r)
	{
		t2 += x - l;
		b[z++] = sum[y++];
	}
	for (int i = l; i <= r; i++) sum[i] = b[i];
}

bool check(int x)
{
	t2 = 0;
	for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + (a[i] <= x ? 1 : -1);
	ms(0, n);
	return t2 >= (t1 / 2 + 1);
}

int main()
{
	int l = inf, r = -inf;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)
	{
		scanf("%d", &a[i]);
		l = min(l, a[i]);
		r = max(r, a[i]);
		t1 += i;
	}
	while (l < r)
	{
		int m = (l + r) >> 1;
		if (check(m)) r = m;
		else l = m + 1;
	}
	printf("%d", r);
	return 0;
}
```