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
/*
ID: jerrywcy
TASK: AtCoder Beginner Contest 107 Problem D
LINK: https://abc107.contest.atcoder.jp/tasks/arc101_b
LANG: C++
*/
#include<bits/stdc++.h>

using namespace std;

long long n,s[100500];

long long solve(long long p) {
    vector<long long> k(2*n+1);
    long long sum=n;
    k[sum]=1;
    long long add=1;
    long long res=0;
    for (int i=1;i<=n;i++) {
        if(s[i]<p) {
            add-=k[sum];
            sum--;
        }
        else {
            sum++;
            add+=k[sum];
        }
        res+=add;
        k[sum]++;
        add++;    
    }
    return res;
}

int main() {
    long long l=INT_MAX,r=INT_MIN,res=INT_MIN;
    cin>>n;
    int i;
    for(i=1;i<=n;i++) {
        cin>>s[i];
        l=min(l,s[i]);
        r=max(r,s[i]);
    }
    if(l==r)
    {
        cout<<l<<endl;
        return 0;
    }
    long long tot=n*(n+1)/2-(n*(n+1)/4+1);
    while(l<=r) {
        long long mid=(l+r)>>1;
        long long ans=solve(mid);
        if(ans<tot)r=mid-1;
        else {
            res=max(mid,res);
            l=mid+1;
        }
    } 
    cout<<res<<endl;  
}