#include<bits/stdc++.h>
#define read(x) scanf("%d",&x)
using namespace std;
typedef long long LL;
const int maxn=5e5+5;
int a[maxn],c[maxn];
int sum[maxn];
int n;
LL tmp;
int lowbit(int x) {
    return x&(-x);
}
int getsum(int x) {
    int sum=0;
    while(x>0) {
        sum+=c[x];
        x-=lowbit(x);
    }
    return sum;
}
void add(int x) {
    while(x<maxn) {
        c[x]+=1;
        x+=lowbit(x);
    }
}
bool check(int x) {
    memset(c,0,sizeof(c));
    sum[0]=0;
    for(int i=1;i<=n;i++) {
        if(a[i]>=x)sum[i]=1;
        else sum[i]=-1;
        sum[i]=sum[i-1]+sum[i];
    }
    for(int i=0;i<=n;i++) {
        sum[i]+=n+1;
    }
    LL ans=0;
    for(int i=0;i<=n;i++) {
        ans+=getsum(sum[i]);
        add(sum[i]);
    }
    if(ans>=tmp)return true;
    else return false;
}
int main() {
    read(n);
    int minn=1e9,maxx=0;
    for(int i=1;i<=n;i++) {
        read(a[i]);
        minn=min(minn,a[i]);
        maxx=max(maxx,a[i]);
    }
    int l=minn,r=maxx;
    int ans=l;
    tmp=1LL*(n+1)*n/4;
    while(l<=r) {
        int mid=(l+r)/2;
        if(check(mid)) {
            l=mid+1;
            ans=mid;
        }
        else r=mid-1;
    }
    printf("%d\n",ans);
}

### 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
#include<bits/stdc++.h>
#define read(x) scanf("%d",&x)
using namespace std;
typedef long long LL;
const int maxn=5e5+5;
int a[maxn],c[maxn];
int sum[maxn];
int n;
LL tmp;
int lowbit(int x) {
    return x&(-x);
}
int getsum(int x) {
    int sum=0;
    while(x>0) {
        sum+=c[x];
        x-=lowbit(x);
    }
    return sum;
}
void add(int x) {
    while(x<maxn) {
        c[x]+=1;
        x+=lowbit(x);
    }
}
bool check(int x) {
    memset(c,0,sizeof(c));
    sum[0]=0;
    for(int i=1;i<=n;i++) {
        if(a[i]>=x)sum[i]=1;
        else sum[i]=-1;
        sum[i]=sum[i-1]+sum[i];
    }
    for(int i=0;i<=n;i++) {
        sum[i]+=n+1;
    }
    LL ans=0;
    for(int i=0;i<=n;i++) {
        ans+=getsum(sum[i]);
        add(sum[i]);
    }
    if(ans>=tmp)return true;
    else return false;
}
int main() {
    read(n);
    int minn=1e9,maxx=0;
    for(int i=1;i<=n;i++) {
        read(a[i]);
        minn=min(minn,a[i]);
        maxx=max(maxx,a[i]);
    }
    int l=minn,r=maxx;
    int ans=l;
    tmp=1LL*(n+1)*n/4;
    while(l<=r) {
        int mid=(l+r)/2;
        if(check(mid)) {
            l=mid+1;
            ans=mid;
        }
        else r=mid-1;
    }
    printf("%d\n",ans);
}

```