#include <bits/stdc++.h>
using namespace std;
int n, a[1000005], b[1000005], sum[1000005];
long long m;
namespace BIT
{
	long long c[5000005];
	inline int lowbit(int x) { return x & (~x + 1); }
	inline void add(int x, long long val)
	{
		for (int i = x; i < 4 * n; i += lowbit(i))	c[i] += val;
	}
	inline long long sum(int x)
	{
		long long res = 0;
		for (int i = x; i; i -= lowbit(i))	res += c[i];
		return res;
	}
}
bool check(int x)
{
	long long cnt = 0;
	sum[0] = 0;
	for (int i = 1; i <= n; ++i)	sum[i] = sum[i - 1] + (a[i] >= x);
	memset(BIT::c, 0, sizeof(BIT::c));
	for (int i = 0; i <= n; ++i)
	{
		sum[i] = (2 * sum[i] - i) + n + 1;
		cnt += BIT::sum(sum[i]);
		BIT::add(sum[i], 1);
	}
	//printf("%d %lld\n", x, cnt);
	return cnt >= m - m / 2;
}
int main()
{
	scanf("%d", &n);
	m = 1ll * n * (n + 1) / 2;
	for (int i = 1; i <= n; ++i)
	{
		scanf("%d", a + i);
		b[i] = a[i];
	}
	sort(b + 1, b + 1 + n);
	int l = 1, r = n, mid, ans;
	while (l <= r)
	{
		mid = (l + r) >> 1;
		if (check(b[mid]))	l = mid + 1, ans = b[mid];
		else r = mid - 1;
	}
	printf("%d\n", ans);
	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;
int n, a[1000005], b[1000005], sum[1000005];
long long m;
namespace BIT
{
	long long c[5000005];
	inline int lowbit(int x) { return x & (~x + 1); }
	inline void add(int x, long long val)
	{
		for (int i = x; i < 4 * n; i += lowbit(i))	c[i] += val;
	}
	inline long long sum(int x)
	{
		long long res = 0;
		for (int i = x; i; i -= lowbit(i))	res += c[i];
		return res;
	}
}
bool check(int x)
{
	long long cnt = 0;
	sum[0] = 0;
	for (int i = 1; i <= n; ++i)	sum[i] = sum[i - 1] + (a[i] >= x);
	memset(BIT::c, 0, sizeof(BIT::c));
	for (int i = 0; i <= n; ++i)
	{
		sum[i] = (2 * sum[i] - i) + n + 1;
		cnt += BIT::sum(sum[i]);
		BIT::add(sum[i], 1);
	}
	//printf("%d %lld\n", x, cnt);
	return cnt >= m - m / 2;
}
int main()
{
	scanf("%d", &n);
	m = 1ll * n * (n + 1) / 2;
	for (int i = 1; i <= n; ++i)
	{
		scanf("%d", a + i);
		b[i] = a[i];
	}
	sort(b + 1, b + 1 + n);
	int l = 1, r = n, mid, ans;
	while (l <= r)
	{
		mid = (l + r) >> 1;
		if (check(b[mid]))	l = mid + 1, ans = b[mid];
		else r = mid - 1;
	}
	printf("%d\n", ans);
	return 0;
}
```