#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 101000;
int n, nums[N], vec[N], sum[N], t[N];
long long merge( int l, int r ){
	if( l == r )
		return 0;
	int mid = (l+r) / 2;
	long long res = merge(l, mid) + merge(mid+1, r);
	int i = l, j = mid+1, k = l;
	while( i <= mid || j <= r ){
		if( j > r || (i <= mid && sum[i] <= sum[j]) )
			t[k++] = sum[i++];
		else
			t[k++] = sum[j++], res += mid - i + 1;
	}
	for( int i = l; i <= r; ++i )
		sum[i] = t[i];
	return res;
}
bool check( int x ){
	for( int i = 1; i <= n; ++i )
		sum[i] = sum[i-1] + (nums[i] < x ? -1 : 1);
	return merge(0, n) * 2 <= (long long)n * (n+1) / 2;
}
int main(){
	scanf("%d", &n);
	for( int i = 1; i <= n; ++i ){
		scanf("%d", &nums[i]);
		vec[i] = nums[i];
	}
	sort(vec+1, vec+1+n);
	int m = unique(vec+1, vec+1+n)-vec;
	int l = 0, r = m-1;
	while( l < r ){
		int mid = (l+r+1) / 2;
		if( check(vec[mid]) )
			l = mid;
		else
			r = mid-1;
	}
	printf("%d\n", vec[r]);
	return 0;
}
### Prompt

In cpp, your task is to solve 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 <cstdio>
#include <algorithm>
using namespace std;
const int N = 101000;
int n, nums[N], vec[N], sum[N], t[N];
long long merge( int l, int r ){
	if( l == r )
		return 0;
	int mid = (l+r) / 2;
	long long res = merge(l, mid) + merge(mid+1, r);
	int i = l, j = mid+1, k = l;
	while( i <= mid || j <= r ){
		if( j > r || (i <= mid && sum[i] <= sum[j]) )
			t[k++] = sum[i++];
		else
			t[k++] = sum[j++], res += mid - i + 1;
	}
	for( int i = l; i <= r; ++i )
		sum[i] = t[i];
	return res;
}
bool check( int x ){
	for( int i = 1; i <= n; ++i )
		sum[i] = sum[i-1] + (nums[i] < x ? -1 : 1);
	return merge(0, n) * 2 <= (long long)n * (n+1) / 2;
}
int main(){
	scanf("%d", &n);
	for( int i = 1; i <= n; ++i ){
		scanf("%d", &nums[i]);
		vec[i] = nums[i];
	}
	sort(vec+1, vec+1+n);
	int m = unique(vec+1, vec+1+n)-vec;
	int l = 0, r = m-1;
	while( l < r ){
		int mid = (l+r+1) / 2;
		if( check(vec[mid]) )
			l = mid;
		else
			r = mid-1;
	}
	printf("%d\n", vec[r]);
	return 0;
}
```