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 re register
#define int long long
using namespace std;
const int maxn=1e5+10;
int n,YZ,a[maxn],tree[maxn<<1],aa[maxn],l,r,mid;
int lowbit(int x){
    return x&(-x);
}
void add(int x){
    while(x<(maxn<<1)){
        ++tree[x];
        x+=lowbit(x);
    }
}
int query(int x){
    int ans=0;
    while(x){
        ans+=tree[x];
        x-=lowbit(x);
    }
    return ans;
}
bool check(int m){
    int cnt=0;
    memset(tree,0,sizeof(tree));
    for(re int i=1;i<=n;++i){
        if(a[i]<=m){
            aa[i]=1;
        }
        else{
            aa[i]=-1;
        }
    }
    aa[0]=0;
    for(re int i=1;i<=n;++i){
        aa[i]+=aa[i-1];
    }
    for(re int i=0;i<=n;++i){
        aa[i]+=maxn;
    }
    add(aa[0]);
    for(re int i=1;i<=n;++i){
        cnt+=query(aa[i]-1);
        add(aa[i]);
    }
    return cnt>=YZ;
}
signed main(){
    scanf("%lld",&n);
    YZ=n*(n+1)/4+1;
    for(re int i=1;i<=n;++i){
        scanf("%lld",&a[i]);
    }
    l=0,r=1e9;
    while(l<r){
        mid=(l+r)>>1;
        if(check(mid)){
            r=mid;
        }
        else{
            l=mid+1;
        }
    }
    printf("%lld\n",l);
    return 0;
}