#include "bits/stdc++.h"
using namespace std;
using ll = long long;
struct BIT {
	ll M;
	std::vector<ll> bit;
	BIT(ll n) {
		init(n);
	}
	void init(ll n) {
		M = n;
		bit.resize(M + 1);
		fill(bit.begin(), bit.end(), 0);
	}
	ll sum(ll i) {
		ll s = 0;
		while (i > 0) {
			s += bit[i];
			i -= i & -i;
		}
		return s;
	}
	void add(ll i, ll x) {
		while (i <= M) {
			bit[i] += x;
			i += i & -i;
		}
	}
};


int main() {
	int N;
	cin >> N;
	vector<int>A(N);
	vector<int>A_Sort(N);
	for (int i = 0; i < N; ++i) {
		cin >> A[i];
		A_Sort[i] = A[i];
	}
	sort(A_Sort.begin(), A_Sort.end());
	int l = 0;
	int r = N;
	while (l + 1 < r) {
		int c = (l + r) / 2;
		vector<int>B(N);
		for (int i = 0; i < N; ++i) {
			B[i] = (A[i] < A_Sort[c]) ? -1 : 1;
		}
		vector<int>S(N);
		for (int i = 0; i < N; ++i) {
			S[i] = (i == 0) ? B[i] : S[i - 1] + B[i];
		}
		long long cnt = 0;
		BIT bit(2 * N + 1);
		bit.add(N, 1);
		for (int i = 0; i < N; ++i) {
			cnt += bit.sum(S[i] + N);
			bit.add(S[i] + N, 1);
		}
		if (cnt >= (long long)N*(N + 1) / 4) {
			l = c;
		}
		else {
			r = c;
		}
	}
	cout << A_Sort[l] << endl;
	return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
using ll = long long;
struct BIT {
	ll M;
	std::vector<ll> bit;
	BIT(ll n) {
		init(n);
	}
	void init(ll n) {
		M = n;
		bit.resize(M + 1);
		fill(bit.begin(), bit.end(), 0);
	}
	ll sum(ll i) {
		ll s = 0;
		while (i > 0) {
			s += bit[i];
			i -= i & -i;
		}
		return s;
	}
	void add(ll i, ll x) {
		while (i <= M) {
			bit[i] += x;
			i += i & -i;
		}
	}
};


int main() {
	int N;
	cin >> N;
	vector<int>A(N);
	vector<int>A_Sort(N);
	for (int i = 0; i < N; ++i) {
		cin >> A[i];
		A_Sort[i] = A[i];
	}
	sort(A_Sort.begin(), A_Sort.end());
	int l = 0;
	int r = N;
	while (l + 1 < r) {
		int c = (l + r) / 2;
		vector<int>B(N);
		for (int i = 0; i < N; ++i) {
			B[i] = (A[i] < A_Sort[c]) ? -1 : 1;
		}
		vector<int>S(N);
		for (int i = 0; i < N; ++i) {
			S[i] = (i == 0) ? B[i] : S[i - 1] + B[i];
		}
		long long cnt = 0;
		BIT bit(2 * N + 1);
		bit.add(N, 1);
		for (int i = 0; i < N; ++i) {
			cnt += bit.sum(S[i] + N);
			bit.add(S[i] + N, 1);
		}
		if (cnt >= (long long)N*(N + 1) / 4) {
			l = c;
		}
		else {
			r = c;
		}
	}
	cout << A_Sort[l] << endl;
	return 0;
}

```