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

struct BIT {
	long long n;
	vector<long long> dat;

	void init(long long n_input) {
		n = n_input + 1;
		dat.resize(n + 1);
		for (int i = 0; i <= n; i++) dat[i] = 0;
	}

	long long sum(long long i) {
		long long s = 0;
		while (i >= 0) {
			s += dat[i];
			i = (i & (i + 1)) - 1;
		}
		return s;
	}

	void add(long long i, long long x) {
		while (i <= n) {
			dat[i] += x;
			i |= i + 1;
		}
	}
};

int main() {
	int i;
	int N;
	cin >> N;
	vector<int> A(N), sortedA(N);
	for (i = 0; i < N; i++) {
		cin >> A[i];
		sortedA[i] = A[i];
	}
	sort(sortedA.begin(), sortedA.end());
	int ub = N;
	int lb = 0;
	BIT bit;

	long long border = (((long long)N * (N + 1) / 2) + 1) / 2;
	while (ub - lb > 1) {
		int mid = (ub + lb) / 2;
		int midn = sortedA[mid];
		bit.init(2 * N + 1);
		int sum = 0;
		bit.add(sum + N, 1);
		long long result = 0;
		for (i = 0; i < N; i++) {
			sum += A[i] >= midn ? 1 : -1;
			result += bit.sum(sum + N);
			bit.add(sum + N, 1);
		}
		if (result >= border) {
			lb = mid;
		}
		else {
			ub = mid;
		}
	}
	cout << sortedA[lb] << endl;
	return 0;
}

### Prompt

Please create a solution in Cpp 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;

struct BIT {
	long long n;
	vector<long long> dat;

	void init(long long n_input) {
		n = n_input + 1;
		dat.resize(n + 1);
		for (int i = 0; i <= n; i++) dat[i] = 0;
	}

	long long sum(long long i) {
		long long s = 0;
		while (i >= 0) {
			s += dat[i];
			i = (i & (i + 1)) - 1;
		}
		return s;
	}

	void add(long long i, long long x) {
		while (i <= n) {
			dat[i] += x;
			i |= i + 1;
		}
	}
};

int main() {
	int i;
	int N;
	cin >> N;
	vector<int> A(N), sortedA(N);
	for (i = 0; i < N; i++) {
		cin >> A[i];
		sortedA[i] = A[i];
	}
	sort(sortedA.begin(), sortedA.end());
	int ub = N;
	int lb = 0;
	BIT bit;

	long long border = (((long long)N * (N + 1) / 2) + 1) / 2;
	while (ub - lb > 1) {
		int mid = (ub + lb) / 2;
		int midn = sortedA[mid];
		bit.init(2 * N + 1);
		int sum = 0;
		bit.add(sum + N, 1);
		long long result = 0;
		for (i = 0; i < N; i++) {
			sum += A[i] >= midn ? 1 : -1;
			result += bit.sum(sum + N);
			bit.add(sum + N, 1);
		}
		if (result >= border) {
			lb = mid;
		}
		else {
			ub = mid;
		}
	}
	cout << sortedA[lb] << endl;
	return 0;
}

```