#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=1e5+5;
ll n,a[N],val[N],prfx[N],bit[2*N+10];
void update(ll x){
    x+=n+2;
    for(;x<=2*n+10;x+=x&(-x))++bit[x];
}
ll query(ll x){
    x+=n+2;
    ll ans = 0;
    for(;x;x-=x&(-x))ans+=bit[x];
    return ans;
}
bool chk(ll val){
    for(ll i=0;i<=2*n+10;i++)bit[i]=0;
    for(ll i=1;i<=n;i++){
        if(a[i]<=val)prfx[i]=1;
        else prfx[i]=-1;
        prfx[i]+=prfx[i-1];
    }
    ll cur = 0 , req = (n*(n+1))/2;
    update(0);
    for(ll i=1;i<=n;i++)cur+=query(prfx[i]-1),update(prfx[i]);
    return cur>=req/2  + 1;
}
int main(){
    ios_base:: sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    cin>>n;
    for(ll i=1;i<=n;i++)cin>>a[i],val[i]=a[i];
    sort(val+1,val+n+1);
    ll k = unique(val+1,val+n+1)-val-1;
    ll l=1,r=k,ans=0;
    while(l<=r){
        ll mid=(l+r)>>1;
        if(chk(val[mid]))ans=val[mid],r=mid-1;
        else l=mid+1;
    }
    cout<<ans;
}
### Prompt

In CPP, your task is to solve the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=1e5+5;
ll n,a[N],val[N],prfx[N],bit[2*N+10];
void update(ll x){
    x+=n+2;
    for(;x<=2*n+10;x+=x&(-x))++bit[x];
}
ll query(ll x){
    x+=n+2;
    ll ans = 0;
    for(;x;x-=x&(-x))ans+=bit[x];
    return ans;
}
bool chk(ll val){
    for(ll i=0;i<=2*n+10;i++)bit[i]=0;
    for(ll i=1;i<=n;i++){
        if(a[i]<=val)prfx[i]=1;
        else prfx[i]=-1;
        prfx[i]+=prfx[i-1];
    }
    ll cur = 0 , req = (n*(n+1))/2;
    update(0);
    for(ll i=1;i<=n;i++)cur+=query(prfx[i]-1),update(prfx[i]);
    return cur>=req/2  + 1;
}
int main(){
    ios_base:: sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    cin>>n;
    for(ll i=1;i<=n;i++)cin>>a[i],val[i]=a[i];
    sort(val+1,val+n+1);
    ll k = unique(val+1,val+n+1)-val-1;
    ll l=1,r=k,ans=0;
    while(l<=r){
        ll mid=(l+r)>>1;
        if(chk(val[mid]))ans=val[mid],r=mid-1;
        else l=mid+1;
    }
    cout<<ans;
}
```