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
#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;
using lint=long long;

template<class T>
class Fenwick_tree{
	vector<T> a;
public:
	Fenwick_tree(int n):a(n){}
	void add(int i,T val){
		for(;i<a.size();i|=i+1) a[i]+=val;
	}
	T sum(int l,int r){
		if(l==0){
			T res{};
			for(r--;r>=0;r=(r&(r+1))-1) res+=a[r];
			return res;
		}
		return sum(0,r)-sum(0,l);
	}
};

int n,a[100000];

// number of pairs (l, r) such that median of a[l..r) >= x
lint solve(int x){
	int b[100000];
	rep(i,n) b[i]=(a[i]>=x?1:-1);

	lint res=0;
	int sum=0;
	Fenwick_tree<int> F(2*n+1);
	F.add(n,1);
	rep(i,n){
		sum+=b[i];
		res+=F.sum(0,n+sum+1);
		F.add(n+sum,1);
	}
	return res;
}

int main(){
	scanf("%d",&n);
	rep(i,n) scanf("%d",a+i);

	lint m=n*(n+1LL)/2;

	int lo=0,hi=1e9+1;
	while(hi-lo>1){
		int mi=(lo+hi)/2;
		if((m+1)/2<=solve(mi)) lo=mi;
		else                   hi=mi;
	}
	printf("%d\n",lo);

	return 0;
}
