#include <bits/stdc++.h>
#define rep(i, a, b) for(int i = int(a); i < int(b); i++)
#define rer(i, a, b) for(int i = int(a) - 1; i >= int(b); i--)

using namespace std;
typedef long long int ll;
typedef pair <int, int> P;

const ll MOD=1000000007;
const int MAX_N=1e5*2;
int bit[MAX_N+10], n;
ll a[100000];
int sum(int i){
    int s=0;
    while(i>0){
        s+=bit[i];
        i-=i&-i;
    }
    return s;
}

void add(int i, int x){
    while(i<=MAX_N+10){
        bit[i]+=x;
        i+=i&-i;
    }
}
bool C(ll x){
    int s[n+1];
    s[0]=1e5+1;
    ll res=0;
    rep(i,0,n){
        if(a[i]<=x){
            s[i+1]=s[i]+1;
        }
        else s[i+1]=s[i]-1;
    }
    rep(i,0,n+1){
        res+=sum(s[i]-1);
        add(s[i],1);
    }
    if(res<ll(n)*(n+1)/4+1) return false;
    else return true;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n;
    rep(i,0,n) cin>>a[i];
    ll ng=0, ok=1e9;
    while(ok-ng>1){
        ll mid=(ng+ok)/2;
        if(C(mid)) ok=mid;
        else ng=mid;
        memset(bit, 0, sizeof(bit));
    }
    cout <<ok<<"\n";
}
### Prompt

Your challenge is to write a Cpp solution to 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>
#define rep(i, a, b) for(int i = int(a); i < int(b); i++)
#define rer(i, a, b) for(int i = int(a) - 1; i >= int(b); i--)

using namespace std;
typedef long long int ll;
typedef pair <int, int> P;

const ll MOD=1000000007;
const int MAX_N=1e5*2;
int bit[MAX_N+10], n;
ll a[100000];
int sum(int i){
    int s=0;
    while(i>0){
        s+=bit[i];
        i-=i&-i;
    }
    return s;
}

void add(int i, int x){
    while(i<=MAX_N+10){
        bit[i]+=x;
        i+=i&-i;
    }
}
bool C(ll x){
    int s[n+1];
    s[0]=1e5+1;
    ll res=0;
    rep(i,0,n){
        if(a[i]<=x){
            s[i+1]=s[i]+1;
        }
        else s[i+1]=s[i]-1;
    }
    rep(i,0,n+1){
        res+=sum(s[i]-1);
        add(s[i],1);
    }
    if(res<ll(n)*(n+1)/4+1) return false;
    else return true;
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n;
    rep(i,0,n) cin>>a[i];
    ll ng=0, ok=1e9;
    while(ok-ng>1){
        ll mid=(ng+ok)/2;
        if(C(mid)) ok=mid;
        else ng=mid;
        memset(bit, 0, sizeof(bit));
    }
    cout <<ok<<"\n";
}
```