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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
typedef long long ll;
using namespace std;
const int N=1e5+10;
int a[N],b[N],c[N],n,tr[N<<1];
int lowbit(int x) {return x&-x;}
void change(int x)
{
    x+=N;
    while(x < (N<<1))
    {
        tr[x]++;
        x+=lowbit(x);
    }
}
int sum(int x)
{
    x+=N;
    int ans=0;
    while(x>=1)
    {
        ans+=tr[x];
        x-=lowbit(x);
    }
    return ans;
}
bool check(int now)
{
    for(int i=1;i<=n;i++) c[i]=c[i-1]+(a[i]<=now?1:-1);
    memset(tr,0,sizeof(tr));
    change(c[0]);
    ll tot=0;
    for(int r=1;r<=n;r++)
    {
        tot+=sum(c[r]-1);
        change(c[r]);
    }
    return tot>=ll(n+1)*n/4+1;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&a[i]);
    int l=1,r=1e9,ans=-1;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if (check(mid)) ans=mid,r=mid-1;
        else l=mid+1;
    }
    printf("%d\n",ans);
}
