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>
using namespace std;
int num[100000+5],a[100000+5],c[100000+5],s[100000+5],t[100000+5],cnt;
long long N,ans;
void merge(int a[],int l,int m,int r){
	int i=l,j=m+1,p=i;
	while(i<=m||j<=r){
		if(i<=m&&(j>r||a[i]<=a[j])) t[p++]=a[i++],ans+=j-m-1;
		else t[p++]=a[j++];
	}
	for(int i=l;i<=r;i++){
		a[i]=t[i];
	}
}
void sort(int a[], int l, int r) {
if (l >= r) return ;
int mid = (l + r) / 2;
sort(a, l, mid); 
sort(a, mid + 1, r); 
merge(a, l, mid, r);
}
bool check(int n){
	ans=0;
	for(int i=1;i<=N;i++){
	     if(a[i]>n) c[i]=1;
	     if(a[i]==n) c[i]=1;
		 if(a[i]<n) c[i]=-1;  	
	}
	s[0]=0;
	for(int i=1;i<=N;i++){
		s[i]=s[i-1]+c[i];
	}
	sort(s,0,N);
	if(ans<=(long long)N*(N+1)/4) return true;
	else return false;
}
int main(){
   cin>>N;
    for(int i = 1 ; i <= N ; ++i) {cin>>a[i];num[i] = a[i];}
    
    sort(num + 1,num + N + 1);
    cnt = unique(num + 1,num + N + 1) - num - 1;
    int L = 1,R = cnt;
    while(L < R) {
        int mid = (L + R + 1) /2;
        if(check(num[mid])) L = mid;
        else R = mid-1;
    }
    cout<<num[R]<<endl;
    return 0;
}
