///http://www.cnblogs.com/henry-1202/p/9537952.html
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

const int maxn=1e5+10;
int a[maxn];
int n;
int c[maxn*15];
int pre[maxn*15];
int lowbit(int x)
{
    return x&-x;
}
void add(int x)
{
    for(int i=x;i<=2*maxn;i+=lowbit(i))c[i]++;
}
LL query(int x)
{
    LL res=0;
    for(int i=x;i>0;i-=lowbit(i))res+=c[i];
    return res;
}

bool check(int x)
{
    memset(c,0,sizeof(c));
    pre[0]=0;
    for(int i=1;i<=n;i++)pre[i]=pre[i-1]+(a[i]>=x?1:-1);
    LL res=0;
    for(int i=0;i<=n;i++){
        res+=query(pre[i]+maxn);
        add(pre[i]+maxn);
    }
    return res>=1LL*n*(n+1)/4;
}

int main()
{
    n=read();
    int L=0,R=0,ans=0;
    for(int i=1;i<=n;i++){
        a[i]=read();
        R=max(R,a[i]);
    }
    while(L<=R){
        int mid=(L+R)>>1;
        if(check(mid)){
            L=mid+1;ans=mid;
        }
        else{
            R=mid-1;
        }
    }
    printf("%d\n",ans);
    return 0;
}

### Prompt

Please formulate 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
///http://www.cnblogs.com/henry-1202/p/9537952.html
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

const int maxn=1e5+10;
int a[maxn];
int n;
int c[maxn*15];
int pre[maxn*15];
int lowbit(int x)
{
    return x&-x;
}
void add(int x)
{
    for(int i=x;i<=2*maxn;i+=lowbit(i))c[i]++;
}
LL query(int x)
{
    LL res=0;
    for(int i=x;i>0;i-=lowbit(i))res+=c[i];
    return res;
}

bool check(int x)
{
    memset(c,0,sizeof(c));
    pre[0]=0;
    for(int i=1;i<=n;i++)pre[i]=pre[i-1]+(a[i]>=x?1:-1);
    LL res=0;
    for(int i=0;i<=n;i++){
        res+=query(pre[i]+maxn);
        add(pre[i]+maxn);
    }
    return res>=1LL*n*(n+1)/4;
}

int main()
{
    n=read();
    int L=0,R=0,ans=0;
    for(int i=1;i<=n;i++){
        a[i]=read();
        R=max(R,a[i]);
    }
    while(L<=R){
        int mid=(L+R)>>1;
        if(check(mid)){
            L=mid+1;ans=mid;
        }
        else{
            R=mid-1;
        }
    }
    printf("%d\n",ans);
    return 0;
}

```