#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

typedef long long LL;

const int MN = 100005;

int N, M;
int a[MN], b[MN];
int c[MN];

int B[MN * 2];
void Add(int i) { for (; i <= N * 2; i += i & -i) ++B[i]; }
int Qur(int i) { int A = 0; for (; i; i -= i & -i) A += B[i]; return A; }

inline bool check(int x) {
	for (int i = 1; i <= N; ++i)
		c[i] = c[i - 1] + (a[i] >= x);
	memset(B, 0, sizeof B);
	LL Sum = 0, NN = (LL)N * (N + 1) / 2;
	for (int i = 1; i <= N; ++i) {
		Add(2 * c[i - 1] - (i - 1) + N);
		Sum += Qur(2 * c[i] - i + N);
	}
	return Sum * 2 >= NN;
}

int main() {
	scanf("%d", &N);
	for (int i = 1; i <= N; ++i)
		scanf("%d", &a[i]), b[i] = a[i];
	sort(b + 1, b + N + 1);
	M = unique(b + 1, b + N + 1) - b - 1;
	int lb = 1, rb = M, mid, ans = 0;
	while (lb <= rb) {
		mid = lb + rb >> 1;
		if (check(b[mid])) ans = mid, lb = mid + 1;
		else rb = mid - 1;
	}
	printf("%d\n", b[ans]);
	return 0;
}
### Prompt

Please formulate 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 <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;

typedef long long LL;

const int MN = 100005;

int N, M;
int a[MN], b[MN];
int c[MN];

int B[MN * 2];
void Add(int i) { for (; i <= N * 2; i += i & -i) ++B[i]; }
int Qur(int i) { int A = 0; for (; i; i -= i & -i) A += B[i]; return A; }

inline bool check(int x) {
	for (int i = 1; i <= N; ++i)
		c[i] = c[i - 1] + (a[i] >= x);
	memset(B, 0, sizeof B);
	LL Sum = 0, NN = (LL)N * (N + 1) / 2;
	for (int i = 1; i <= N; ++i) {
		Add(2 * c[i - 1] - (i - 1) + N);
		Sum += Qur(2 * c[i] - i + N);
	}
	return Sum * 2 >= NN;
}

int main() {
	scanf("%d", &N);
	for (int i = 1; i <= N; ++i)
		scanf("%d", &a[i]), b[i] = a[i];
	sort(b + 1, b + N + 1);
	M = unique(b + 1, b + N + 1) - b - 1;
	int lb = 1, rb = M, mid, ans = 0;
	while (lb <= rb) {
		mid = lb + rb >> 1;
		if (check(b[mid])) ans = mid, lb = mid + 1;
		else rb = mid - 1;
	}
	printf("%d\n", b[ans]);
	return 0;
}
```