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

const int maxn = 1e5+5;
int n, a[maxn];
int pre[maxn];

int bit[3*maxn];
void add(int i, int x) {
    i += maxn;
    for (; i < 3*maxn; i += i & (-i))
        bit[i] += x;
}

int sum(int i) {
    i += maxn;
    int r = 0;
    for (; i; i -= i & (-i)) {
        r += bit[i];
    }
    return r;
}

ll cnt(int x) {
    for (int i = 1; i <= n; i++) {
        pre[i] = a[i] < x ? -1 : 1;
        pre[i] += pre[i-1];
    }
    memset(bit,0,sizeof bit);
    ll res = 0;
    add(0,1);
    for (int i = 1; i <= n; i++) {
        res += sum(2*maxn)-sum(pre[i]);    
        add(pre[i],1);
    }
    return res;
}

int main() {
    ios_base::sync_with_stdio(false); cin.tie(NULL);
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    ll num = 1LL*n*(n+1)/4;
    int lo = 0, hi = 1e9+9;
    while (lo + 1 < hi) {
        int mid = (lo+hi)/2;
        //cout << mid << ": " << cnt(mid) << '\n';
        if (cnt(mid) <= num) lo = mid;
        else hi = mid;
    }
    cout << lo << '\n';
    /*
    cout << '\n';
    for (int i = hi-5; i <= hi+15; i++) {
        cout << i << ": " << cnt(i) << '\n';
    }
    */
}

### Prompt

Your task is to create 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>
using namespace std;
using ll = long long;

const int maxn = 1e5+5;
int n, a[maxn];
int pre[maxn];

int bit[3*maxn];
void add(int i, int x) {
    i += maxn;
    for (; i < 3*maxn; i += i & (-i))
        bit[i] += x;
}

int sum(int i) {
    i += maxn;
    int r = 0;
    for (; i; i -= i & (-i)) {
        r += bit[i];
    }
    return r;
}

ll cnt(int x) {
    for (int i = 1; i <= n; i++) {
        pre[i] = a[i] < x ? -1 : 1;
        pre[i] += pre[i-1];
    }
    memset(bit,0,sizeof bit);
    ll res = 0;
    add(0,1);
    for (int i = 1; i <= n; i++) {
        res += sum(2*maxn)-sum(pre[i]);    
        add(pre[i],1);
    }
    return res;
}

int main() {
    ios_base::sync_with_stdio(false); cin.tie(NULL);
    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
    }
    ll num = 1LL*n*(n+1)/4;
    int lo = 0, hi = 1e9+9;
    while (lo + 1 < hi) {
        int mid = (lo+hi)/2;
        //cout << mid << ": " << cnt(mid) << '\n';
        if (cnt(mid) <= num) lo = mid;
        else hi = mid;
    }
    cout << lo << '\n';
    /*
    cout << '\n';
    for (int i = hi-5; i <= hi+15; i++) {
        cout << i << ": " << cnt(i) << '\n';
    }
    */
}

```