#include <iostream>
#include <cstdio>
#define N 100005
using namespace std;

typedef long long ll;
ll n, ba, a[N], tr[N * 8];

ll se1(ll p, ll q) {
	ll re = 0;
	while (p < q) {
		if (p % 2 == 1) re += tr[p], p++;
		if (q % 2 == 0) re += tr[q], q--;
		p /= 2; q /= 2;
	}
	if (p == q) re += tr[p];
	return re;
}

void se2(ll p) {
	for (; p; p /= 2) {
		tr[p] = tr[p * 2] + tr[p * 2 + 1];
	}
}

bool f(ll p) {
	ll i, j = ba + n - 1, s = 0;
	for (i = 0; i < ba * 2; i++) tr[i] = 0;
	for (i = 0; i < n; i++) {
		tr[j]++;
		se2(j / 2);
		if (a[i] < p) j--;
		else j++;
		s += se1(ba, j);
	}
	return s >= (n * (n + 1) / 2 + 1) / 2;
}

int main()
{
	ll i, l, r, m;
	cin >> n;
	for (i = 0; i < n; i++) {
		scanf("%lld", &a[i]);
	}
	for (ba = 1; ba < n * 2; ba *= 2);
	l = 0; r = 1e9;
	while (l < r) {
		m = (l + r + 1) / 2;
		if (f(m)) l = m;
		else r = m - 1;
	}
    cout << l << endl;
    return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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 <cstdio>
#define N 100005
using namespace std;

typedef long long ll;
ll n, ba, a[N], tr[N * 8];

ll se1(ll p, ll q) {
	ll re = 0;
	while (p < q) {
		if (p % 2 == 1) re += tr[p], p++;
		if (q % 2 == 0) re += tr[q], q--;
		p /= 2; q /= 2;
	}
	if (p == q) re += tr[p];
	return re;
}

void se2(ll p) {
	for (; p; p /= 2) {
		tr[p] = tr[p * 2] + tr[p * 2 + 1];
	}
}

bool f(ll p) {
	ll i, j = ba + n - 1, s = 0;
	for (i = 0; i < ba * 2; i++) tr[i] = 0;
	for (i = 0; i < n; i++) {
		tr[j]++;
		se2(j / 2);
		if (a[i] < p) j--;
		else j++;
		s += se1(ba, j);
	}
	return s >= (n * (n + 1) / 2 + 1) / 2;
}

int main()
{
	ll i, l, r, m;
	cin >> n;
	for (i = 0; i < n; i++) {
		scanf("%lld", &a[i]);
	}
	for (ba = 1; ba < n * 2; ba *= 2);
	l = 0; r = 1e9;
	while (l < r) {
		m = (l + r + 1) / 2;
		if (f(m)) l = m;
		else r = m - 1;
	}
    cout << l << endl;
    return 0;
}
```