// author: Saman Mahdanian
#include <iostream>
#include <cstring>

#define int long long

using namespace std;
const int N = 1e5 + 10;

int n, a[N], _cnt[2 * N];
long long get_le (int);

int32_t main() {
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> a[i];

	long long g  = 2e9;
	long long le = 0;

	while (g - le > 1) {
		int med = (g + le) / 2;
		if (get_le(med) * 2 > (n * (n + 1)) / 2)
			g = med;
		else
			le = med;
	}

	cout << g << endl;
}

long long get_le (int x) { 
	memset (_cnt, 0, sizeof _cnt);
	int* cnt = &_cnt[N];
	cnt[0] = 1;
	int cur = 0, p = 0;
	long long res = 0;

	for (int i = 0; i < n; i++) {
		int val = a[i] <= x? +1 : -1;
		if (val == 1)
			cur += cnt[p];
		else
			cur -= cnt[p - 1];

		cur = max(cur, 0LL);
		res += cur;
		p += val;
		cnt[p]++;
	}
	return res;
} 

// SamMHD :: 05/12/2018 15:52:56 :: Another Nemidoonam-Che-Kooftie...
### Prompt

Please formulate 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
// author: Saman Mahdanian
#include <iostream>
#include <cstring>

#define int long long

using namespace std;
const int N = 1e5 + 10;

int n, a[N], _cnt[2 * N];
long long get_le (int);

int32_t main() {
	cin >> n;
	for (int i = 0; i < n; i++)
		cin >> a[i];

	long long g  = 2e9;
	long long le = 0;

	while (g - le > 1) {
		int med = (g + le) / 2;
		if (get_le(med) * 2 > (n * (n + 1)) / 2)
			g = med;
		else
			le = med;
	}

	cout << g << endl;
}

long long get_le (int x) { 
	memset (_cnt, 0, sizeof _cnt);
	int* cnt = &_cnt[N];
	cnt[0] = 1;
	int cur = 0, p = 0;
	long long res = 0;

	for (int i = 0; i < n; i++) {
		int val = a[i] <= x? +1 : -1;
		if (val == 1)
			cur += cnt[p];
		else
			cur -= cnt[p - 1];

		cur = max(cur, 0LL);
		res += cur;
		p += val;
		cnt[p]++;
	}
	return res;
} 

// SamMHD :: 05/12/2018 15:52:56 :: Another Nemidoonam-Che-Kooftie...
```