#include <iostream>
#include <vector>
using namespace std;

#include <bits/stdc++.h>
#include<chrono>
#define ll long long
using namespace std;
vector<ll>bit;
ll MOD = 1e9 + 7;
ll A, B, H, N, M, L, K, W, X, Y, Z;
ll ans = 0;

void add(ll a, ll w) {
	for (int x = a; x < 2 * N; x += x & -x) {
		bit[x] += w;
	}
}

int sum(ll a) {
	int res = 0;
	for (int x = a; x > 0; x -= x & -x)res += bit[x];
	return res;
}

bool isOK(int mid, vector<ll>a) {
	vector<ll>b(a.size() + 1);
	vector<ll>csum(a.size() + 1);
	csum[0] = 0;
	for (int i = 1; i < N + 1; i++) {
		if (a[i] >= mid)b[i] = 1;
		else b[i] = -1;
		csum[i] = csum[i - 1] + b[i];
	}
	ll invN = 0;
	add(csum[0] + N, 1);
	for (int i = 1; i < N + 1; i++) {
		invN += sum(csum[i] + N);
		add(csum[i] + N, 1);
	}
	return invN >= N * (N + 1) / 4;
}

int main() {
	cin >> N;
	vector<ll>a(N + 1), b(N + 1), csum(N + 2), copy(3 * N, 0);
	ll r = 0;
	ll l = 0;
	for (int i = 1; i < N + 1; i++) {
		cin >> a[i];
		r = max(r, a[i]);
	}
	r++;
	while (r - l > 1) {
		bit = copy;
		ll mid = (r + l) / 2;
		if (isOK(mid, a))l = mid;
		else r = mid;
	}
	cout << l << endl;
}
### 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 <iostream>
#include <vector>
using namespace std;

#include <bits/stdc++.h>
#include<chrono>
#define ll long long
using namespace std;
vector<ll>bit;
ll MOD = 1e9 + 7;
ll A, B, H, N, M, L, K, W, X, Y, Z;
ll ans = 0;

void add(ll a, ll w) {
	for (int x = a; x < 2 * N; x += x & -x) {
		bit[x] += w;
	}
}

int sum(ll a) {
	int res = 0;
	for (int x = a; x > 0; x -= x & -x)res += bit[x];
	return res;
}

bool isOK(int mid, vector<ll>a) {
	vector<ll>b(a.size() + 1);
	vector<ll>csum(a.size() + 1);
	csum[0] = 0;
	for (int i = 1; i < N + 1; i++) {
		if (a[i] >= mid)b[i] = 1;
		else b[i] = -1;
		csum[i] = csum[i - 1] + b[i];
	}
	ll invN = 0;
	add(csum[0] + N, 1);
	for (int i = 1; i < N + 1; i++) {
		invN += sum(csum[i] + N);
		add(csum[i] + N, 1);
	}
	return invN >= N * (N + 1) / 4;
}

int main() {
	cin >> N;
	vector<ll>a(N + 1), b(N + 1), csum(N + 2), copy(3 * N, 0);
	ll r = 0;
	ll l = 0;
	for (int i = 1; i < N + 1; i++) {
		cin >> a[i];
		r = max(r, a[i]);
	}
	r++;
	while (r - l > 1) {
		bit = copy;
		ll mid = (r + l) / 2;
		if (isOK(mid, a))l = mid;
		else r = mid;
	}
	cout << l << endl;
}
```