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

const long long MAX_N = 1e5 + 5e4 + 5, inf = 1e9 + 7;
long long n;
long long Arr[MAX_N];
long long P[MAX_N];
long long psum[MAX_N];
long long cnt[MAX_N];
long long dp[MAX_N];

void process(long long d) {
	memset(cnt, 0, sizeof cnt);
	memset(dp, 0, sizeof dp);
	for (long long i = 1; i <= n; i++) {
		P[i] = (d <= Arr[i]? +1: -1);
		psum[i] = P[i] + psum[i - 1];
	}
}

long long calc(long long d) {
	long long res = 0;
	cnt[50000] = 1;
	
	for (long long i = 1; i <= n; i++) {
		dp[i] = dp[i - 1];
		if (P[i] == +1)
			dp[i] += cnt[50000 + psum[i - 1]];
		else
			dp[i] -= cnt[50000 + psum[i]];
			
		res += dp[i] + cnt[50000+ psum[i]];
		cnt[50000+ psum[i]]++;
	}
	
	return res;
}

int main() {
	scanf("%d", &n);
	for (long long i = 1; i <= n; i++)
		scanf("%d", &Arr[i]);
	
	long long s = 1, e = inf;
	while (e - s > 1) {
		long long mid = e + s >> 1;
		
		process(mid);
		if (calc(mid) < ((((n * (n + 1)) >> 1) + 1) >> 1))
			e = mid;
		else
			s = mid;
	}
	
	printf("%d\n", s);
	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;

const long long MAX_N = 1e5 + 5e4 + 5, inf = 1e9 + 7;
long long n;
long long Arr[MAX_N];
long long P[MAX_N];
long long psum[MAX_N];
long long cnt[MAX_N];
long long dp[MAX_N];

void process(long long d) {
	memset(cnt, 0, sizeof cnt);
	memset(dp, 0, sizeof dp);
	for (long long i = 1; i <= n; i++) {
		P[i] = (d <= Arr[i]? +1: -1);
		psum[i] = P[i] + psum[i - 1];
	}
}

long long calc(long long d) {
	long long res = 0;
	cnt[50000] = 1;
	
	for (long long i = 1; i <= n; i++) {
		dp[i] = dp[i - 1];
		if (P[i] == +1)
			dp[i] += cnt[50000 + psum[i - 1]];
		else
			dp[i] -= cnt[50000 + psum[i]];
			
		res += dp[i] + cnt[50000+ psum[i]];
		cnt[50000+ psum[i]]++;
	}
	
	return res;
}

int main() {
	scanf("%d", &n);
	for (long long i = 1; i <= n; i++)
		scanf("%d", &Arr[i]);
	
	long long s = 1, e = inf;
	while (e - s > 1) {
		long long mid = e + s >> 1;
		
		process(mid);
		if (calc(mid) < ((((n * (n + 1)) >> 1) + 1) >> 1))
			e = mid;
		else
			s = mid;
	}
	
	printf("%d\n", s);
	return 0;
}
```