#include <bits/stdc++.h>
using namespace std;

const int N = 1e5 + 5;
int n;
int a[N], b[N];
int p[N];
int c[2 * N + 5];

int lowbit(int k) {return k & (-k);}

void Add(int k)
{
    for (int i = k; i <= 2 * N; i += lowbit(i))
        c[i]++;
}

long long Find(int k)
{
    long long sum = 0;
    for (int i = k; i != 0; i -= lowbit(i))
        sum += c[i];
    return sum;
}

long long Calc(int num)
{
    long long ans = 0;
    int sum = 0;
    memset(c, 0, sizeof(c));
    for (int i = 1; i <= n; i++)
    {
        if (a[i] >= num)
        {
            Add(-sum + N);
            sum++;
        }
        else
        {
            Add(-sum + N);
            sum--;
        }
        ans += Find(-sum - 1 + N);
    }
    return ans;
}

int main()
{
    scanf ("%d",&n);
    for (int i = 1; i <= n; i++)
    {
        scanf ("%d",&a[i]);
        b[i] = a[i];
    }
    sort(b + 1, b + 1 + n);
    int l = 1, r = n;
    while (l < r)
    {
        int mid = (l + r + 1) / 2;
        if (Calc(b[mid]) > 1ll * n * (n + 1) / 2 / 2) r = mid - 1;
        else l = mid;
    }
    printf ("%d",b[l]);
}

### Prompt

In CPP, your task is to solve 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>
using namespace std;

const int N = 1e5 + 5;
int n;
int a[N], b[N];
int p[N];
int c[2 * N + 5];

int lowbit(int k) {return k & (-k);}

void Add(int k)
{
    for (int i = k; i <= 2 * N; i += lowbit(i))
        c[i]++;
}

long long Find(int k)
{
    long long sum = 0;
    for (int i = k; i != 0; i -= lowbit(i))
        sum += c[i];
    return sum;
}

long long Calc(int num)
{
    long long ans = 0;
    int sum = 0;
    memset(c, 0, sizeof(c));
    for (int i = 1; i <= n; i++)
    {
        if (a[i] >= num)
        {
            Add(-sum + N);
            sum++;
        }
        else
        {
            Add(-sum + N);
            sum--;
        }
        ans += Find(-sum - 1 + N);
    }
    return ans;
}

int main()
{
    scanf ("%d",&n);
    for (int i = 1; i <= n; i++)
    {
        scanf ("%d",&a[i]);
        b[i] = a[i];
    }
    sort(b + 1, b + 1 + n);
    int l = 1, r = n;
    while (l < r)
    {
        int mid = (l + r + 1) / 2;
        if (Calc(b[mid]) > 1ll * n * (n + 1) / 2 / 2) r = mid - 1;
        else l = mid;
    }
    printf ("%d",b[l]);
}

```