#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
int a[100005];
vector<int> v;
int n;

LL pw[100005], idx[100005];
LL val[100005], sum[100005];
LL diff[400005];
int b = 200000;
LL cal(LL x){
	LL res = 0, ans = 0;
	for(int i=1;i<=n;i++){
		if(a[i] <= x) val[i] = 1;
		else val[i] = -1;
		sum[i] = sum[i-1] + val[i];
		diff[sum[i]+b]++;
		if(sum[i] > 0) res++;
	}
	int cnt = 1;
	for(int i=1;i<=n;i++){
		ans += res;

		if(a[i] <= x) res--;
		diff[sum[i]+b]--;

		if(a[i] <= x){
			res -= diff[cnt+b];
			cnt += val[i];
		}
		else{
			cnt += val[i];
			res += diff[cnt+b];
		}
	}
	return ans;
}

int main(){
	scanf("%d", &n);
	LL m = (LL)n*(n+1)/2;
	LL h = m/2+1;

	for(int i=1;i<=n;i++){
		scanf("%d", &a[i]);
		v.push_back(a[i]);
	}
	sort(v.begin(), v.end());
	v.erase(unique(v.begin(), v.end()), v.end());
	int lo = 0, hi = v.size()-1, ans = -1;
	while(lo <= hi){
		int mid = (lo+hi)/2;
		int val = v[mid];
		if(cal(val) >= h) ans = val, hi = mid-1;
		else lo = mid+1;
	}
	printf("%d\n", ans);
}
### 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;
typedef long long LL;
int a[100005];
vector<int> v;
int n;

LL pw[100005], idx[100005];
LL val[100005], sum[100005];
LL diff[400005];
int b = 200000;
LL cal(LL x){
	LL res = 0, ans = 0;
	for(int i=1;i<=n;i++){
		if(a[i] <= x) val[i] = 1;
		else val[i] = -1;
		sum[i] = sum[i-1] + val[i];
		diff[sum[i]+b]++;
		if(sum[i] > 0) res++;
	}
	int cnt = 1;
	for(int i=1;i<=n;i++){
		ans += res;

		if(a[i] <= x) res--;
		diff[sum[i]+b]--;

		if(a[i] <= x){
			res -= diff[cnt+b];
			cnt += val[i];
		}
		else{
			cnt += val[i];
			res += diff[cnt+b];
		}
	}
	return ans;
}

int main(){
	scanf("%d", &n);
	LL m = (LL)n*(n+1)/2;
	LL h = m/2+1;

	for(int i=1;i<=n;i++){
		scanf("%d", &a[i]);
		v.push_back(a[i]);
	}
	sort(v.begin(), v.end());
	v.erase(unique(v.begin(), v.end()), v.end());
	int lo = 0, hi = v.size()-1, ans = -1;
	while(lo <= hi){
		int mid = (lo+hi)/2;
		int val = v[mid];
		if(cal(val) >= h) ans = val, hi = mid-1;
		else lo = mid+1;
	}
	printf("%d\n", ans);
}
```