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>
const int N=1e5+5;
typedef long long ll;
int n,a[N],b[N],c[N],d[N];
int bit[N<<1];ll tot=0;
void init() {
    for(int i=1;i<N<<1;i++) bit[i]=0;
}
void add(int x,int v) {
    for(;x<N<<1;x+=x&-x) bit[x]+=v;
}
int sum(int x) {
    int res=0;
    for(;x;x-=x&-x) res+=bit[x];
    return res;
}
bool check(int x) {
    init();
    for(int i=1;i<=n;i++) {
        if(a[i]<x) b[i]=-1;
        else b[i]=1;
        c[i]=c[i-1]+b[i];
        add(c[i]+N,1);
    }
    ll ans=0;int now=N;
    for(int i=1;i<=n;i++) {
        ans+=sum(now-1);
        now+=b[i];add(c[i]+N,-1);
    }
    return tot-ans>=ans;
}
int main() {
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",a+i),d[i]=a[i];
    int l=1,r=1e9+7,mid;tot=1LL*(n+1)*n>>1;tot=tot;
    std::sort(d+1,d+n+1);int tail=std::unique(d+1,d+n+1)-d;
    r=tail;
    while(r-l>1) {
        mid=(l+r)>>1;
        if(check(d[mid])) l=mid;
        else r=mid;
    }
    printf("%d\n",d[l]);
    return 0;
}
