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

const int N = 110000;

int a[N], sum[N];
int BIT[3 * N];

int main() {
	
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	
	int n; cin >> n;
	for (int i = 1; i <= n; i++) cin >> a[i];
	
	int st = 0, en = 1e9, mid, ans = en;
	
	auto add = [&](int u, int v) {
		for (int i = u; i <= 3 * (n + 1); i += i & -i) BIT[i] += v;
	};
	
	auto get = [&](int u) {
		int ret = 0;
		for (int i = u; i > 0; i -= i & -i) ret += BIT[i];
		return ret;
	};
	
	auto can = [&](int mid) {
		sum[0] = 0;
		for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + 2 * (a[i] <= mid);
		for (int i = 0; i <= 3 * (n + 1); i++) BIT[i] = 0;
		long long cnt = 0;
		add(0 + n + 1, 1);
		for (int i = 1; i <= n; i++) {
			cnt += get(sum[i] - i + n);
			add(sum[i] - i + n + 1, 1);
		}
		return cnt * 2 > (long long)n * (n + 1) / 2;
	};
	
	while (st <= en) {
		mid = st + en >> 1;
		if (can(mid)) ans = mid, en = mid - 1;
		else st = mid + 1;
	}
	cout << ans << "\n";
	
	return 0;
}
### 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
#include <bits/stdc++.h>
using namespace std;

const int N = 110000;

int a[N], sum[N];
int BIT[3 * N];

int main() {
	
	ios::sync_with_stdio(0);
	cin.tie(0), cout.tie(0);
	
	int n; cin >> n;
	for (int i = 1; i <= n; i++) cin >> a[i];
	
	int st = 0, en = 1e9, mid, ans = en;
	
	auto add = [&](int u, int v) {
		for (int i = u; i <= 3 * (n + 1); i += i & -i) BIT[i] += v;
	};
	
	auto get = [&](int u) {
		int ret = 0;
		for (int i = u; i > 0; i -= i & -i) ret += BIT[i];
		return ret;
	};
	
	auto can = [&](int mid) {
		sum[0] = 0;
		for (int i = 1; i <= n; i++) sum[i] = sum[i - 1] + 2 * (a[i] <= mid);
		for (int i = 0; i <= 3 * (n + 1); i++) BIT[i] = 0;
		long long cnt = 0;
		add(0 + n + 1, 1);
		for (int i = 1; i <= n; i++) {
			cnt += get(sum[i] - i + n);
			add(sum[i] - i + n + 1, 1);
		}
		return cnt * 2 > (long long)n * (n + 1) / 2;
	};
	
	while (st <= en) {
		mid = st + en >> 1;
		if (can(mid)) ans = mid, en = mid - 1;
		else st = mid + 1;
	}
	cout << ans << "\n";
	
	return 0;
}
```