#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define pb push_back
const int N = 5e5;
int a[N];
ll n;
ll suf[N];
ll aux[N];
ll bit[N];
void add(int idx, int val){
    for(int i=idx;i>0;i-=(i&-i)){
        bit[i]+=val;
    }
}
ll sum(int idx){
    ll ret = 0;
    for(int i=idx;i<N;i+=(i&-i)){
        ret+=bit[i];
    }
    return ret;
}
bool ch(ll x){
    for(int i=0;i<N;i++)bit[i] = 0;
    ll k = n*(n+1ll)/2;k = k/2+1;
    suf[n+1] = 0;
    aux[n+1] = 0;
    int trt = 1;
    for(int i=n;i>=1;i--){
        suf[i] = suf[i+1];
        if(a[i]<=x)suf[i]++;
        aux[i] = 2*suf[i]-trt++;
    }
    ll ans = 0;
    int zero = n+100;
    for(int i=1;i<=n+1;i++){
        ans+=sum(zero+aux[i]+1);
        add(zero+aux[i],1);
    }
    if(ans<k)return true;
    return false;
}
int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    ll st = 0;
    ll en = 1e18;
    while(st<=en){
        ll mid = (st+en)/2ll;
        if(ch(mid)){
            st = mid+1;
        }
        else{
            en = mid-1;
        }
    }
    cout<<st<<endl;
    return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
#define pb push_back
const int N = 5e5;
int a[N];
ll n;
ll suf[N];
ll aux[N];
ll bit[N];
void add(int idx, int val){
    for(int i=idx;i>0;i-=(i&-i)){
        bit[i]+=val;
    }
}
ll sum(int idx){
    ll ret = 0;
    for(int i=idx;i<N;i+=(i&-i)){
        ret+=bit[i];
    }
    return ret;
}
bool ch(ll x){
    for(int i=0;i<N;i++)bit[i] = 0;
    ll k = n*(n+1ll)/2;k = k/2+1;
    suf[n+1] = 0;
    aux[n+1] = 0;
    int trt = 1;
    for(int i=n;i>=1;i--){
        suf[i] = suf[i+1];
        if(a[i]<=x)suf[i]++;
        aux[i] = 2*suf[i]-trt++;
    }
    ll ans = 0;
    int zero = n+100;
    for(int i=1;i<=n+1;i++){
        ans+=sum(zero+aux[i]+1);
        add(zero+aux[i],1);
    }
    if(ans<k)return true;
    return false;
}
int main(){
    ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    cin>>n;
    for(int i=1;i<=n;i++)cin>>a[i];
    ll st = 0;
    ll en = 1e18;
    while(st<=en){
        ll mid = (st+en)/2ll;
        if(ch(mid)){
            st = mid+1;
        }
        else{
            en = mid-1;
        }
    }
    cout<<st<<endl;
    return 0;
}

```