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

static long long N = 200000, M, L = -1, R = 1000000001, A[100010] = {}, BIT[200010] = {};
vector<long long> V;

void Add(int X, long long Y) {
	for (int i = X; i <= N + 1; i += i & -i) BIT[i] += Y;
}

long long Sum(int X) {
	long long Ret = 0;
	for (int i = X; i > 0; i -= i & -i) Ret += BIT[i];
	return Ret;
}

int main() {
	cin >> M;
	for (int i = 0; i < M; i++) cin >> A[i];
	while (R - L > 1) {
		long long MID = (L + R) / 2, COUNT = 0;
		static long long X[100010] = {};
		X[0] = 100001;
		for (int i = 0; i < M; i++) {
			if (A[i] < MID) X[i + 1] = X[i] - 1;
			else X[i + 1] = X[i] + 1;
		}
		for (int i = 0; i <= 200000; i++) BIT[i] = 0;
		for (int i = 0; i <= M; i++) {
			COUNT += Sum(X[i]);
			Add(X[i], 1);
		}
		if (COUNT < M * (M + 1) / 2 - COUNT) R = MID;
		else L = MID;
 	}
	cout << L << endl;
}
### Prompt

Your challenge is to write 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;

static long long N = 200000, M, L = -1, R = 1000000001, A[100010] = {}, BIT[200010] = {};
vector<long long> V;

void Add(int X, long long Y) {
	for (int i = X; i <= N + 1; i += i & -i) BIT[i] += Y;
}

long long Sum(int X) {
	long long Ret = 0;
	for (int i = X; i > 0; i -= i & -i) Ret += BIT[i];
	return Ret;
}

int main() {
	cin >> M;
	for (int i = 0; i < M; i++) cin >> A[i];
	while (R - L > 1) {
		long long MID = (L + R) / 2, COUNT = 0;
		static long long X[100010] = {};
		X[0] = 100001;
		for (int i = 0; i < M; i++) {
			if (A[i] < MID) X[i + 1] = X[i] - 1;
			else X[i + 1] = X[i] + 1;
		}
		for (int i = 0; i <= 200000; i++) BIT[i] = 0;
		for (int i = 0; i <= M; i++) {
			COUNT += Sum(X[i]);
			Add(X[i], 1);
		}
		if (COUNT < M * (M + 1) / 2 - COUNT) R = MID;
		else L = MID;
 	}
	cout << L << endl;
}
```