#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1e5+10;
int arr[maxn];
int mx=0,n;
int b[maxn*2];
int bj[maxn];
int lowbit(int x){return x&(-x);}
void add(int x){for(int i=x;i<maxn*2;i+=lowbit(i))b[i]+=1;}
int query(int x){int ans=0;for(int i=x;i>0;i-=lowbit(i))ans+=b[i];return ans;}
bool check(int x){
    memset(b,0,sizeof(b));
    for(int i=1;i<=n;i++){
        if(arr[i]>=x)bj[i]=1;
        else bj[i]=-1;
    }
    for(int i=1;i<=n;i++)bj[i]+=bj[i-1];
    ll ans=0;
    for(int i=0;i<=n;i++){
        ans+=query(bj[i]+100001);
        add(bj[i]+100001);
    }
    if(ans>=1ll*(n)*(n+1)/4)return 1;
    else return 0;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&arr[i]),mx=max(mx,arr[i]);
    int l=1,r=mx,ans;
    while(l<=r){
        int mid=(l+r)/2;
        //cout<<mid<<endl;
        if(check(mid))l=mid+1,ans=mid;
        else r=mid-1;
    }
    printf("%d\n",ans);
}

### 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 ll long long
using namespace std;
const int maxn=1e5+10;
int arr[maxn];
int mx=0,n;
int b[maxn*2];
int bj[maxn];
int lowbit(int x){return x&(-x);}
void add(int x){for(int i=x;i<maxn*2;i+=lowbit(i))b[i]+=1;}
int query(int x){int ans=0;for(int i=x;i>0;i-=lowbit(i))ans+=b[i];return ans;}
bool check(int x){
    memset(b,0,sizeof(b));
    for(int i=1;i<=n;i++){
        if(arr[i]>=x)bj[i]=1;
        else bj[i]=-1;
    }
    for(int i=1;i<=n;i++)bj[i]+=bj[i-1];
    ll ans=0;
    for(int i=0;i<=n;i++){
        ans+=query(bj[i]+100001);
        add(bj[i]+100001);
    }
    if(ans>=1ll*(n)*(n+1)/4)return 1;
    else return 0;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%d",&arr[i]),mx=max(mx,arr[i]);
    int l=1,r=mx,ans;
    while(l<=r){
        int mid=(l+r)/2;
        //cout<<mid<<endl;
        if(check(mid))l=mid+1,ans=mid;
        else r=mid-1;
    }
    printf("%d\n",ans);
}

```