#include<bits/stdc++.h>
using namespace std;
int a[110000],sum[110000];
int b[330000];
int main() {
	int n; cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i];
	int st=0,en=1e9,mid,ans=en;
	auto add=[&](int u,int v) {
		for(int i=u;i<=3*(n+1);i+=i&-i) b[i]+=v;
	};
	auto get=[&](int u) {
		int ret=0;
		for(int i=u;i>0;i-=i&-i) ret+=b[i];
		return ret;
	};
	auto can=[&](int mid) {
		sum[0]=0;
		for(int i=1;i<=n;i++) sum[i]=sum[i-1]+2*(a[i]<=mid);
		for(int i=0;i<=3*(n+1);i++) b[i]=0;
		long long cnt=0;
		add(0+n+1,1);
		for (int i=1;i<=n;i++){
			cnt+=get(sum[i]-i+n);
			add(sum[i]-i+n+1,1);
		}
		return cnt*2>(long long)n*(n+1)/2;
	};
	while(st<=en) {
		mid=st+en>>1;
		if (can(mid)) ans=mid,en=mid-1;
		else st=mid+1;
	}
	cout<<ans;
	return 0;
}

### Prompt

Develop a solution in Cpp 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<bits/stdc++.h>
using namespace std;
int a[110000],sum[110000];
int b[330000];
int main() {
	int n; cin>>n;
	for(int i=1;i<=n;i++) cin>>a[i];
	int st=0,en=1e9,mid,ans=en;
	auto add=[&](int u,int v) {
		for(int i=u;i<=3*(n+1);i+=i&-i) b[i]+=v;
	};
	auto get=[&](int u) {
		int ret=0;
		for(int i=u;i>0;i-=i&-i) ret+=b[i];
		return ret;
	};
	auto can=[&](int mid) {
		sum[0]=0;
		for(int i=1;i<=n;i++) sum[i]=sum[i-1]+2*(a[i]<=mid);
		for(int i=0;i<=3*(n+1);i++) b[i]=0;
		long long cnt=0;
		add(0+n+1,1);
		for (int i=1;i<=n;i++){
			cnt+=get(sum[i]-i+n);
			add(sum[i]-i+n+1,1);
		}
		return cnt*2>(long long)n*(n+1)/2;
	};
	while(st<=en) {
		mid=st+en>>1;
		if (can(mid)) ans=mid,en=mid-1;
		else st=mid+1;
	}
	cout<<ans;
	return 0;
}

```