#include "bits/stdc++.h"
using namespace std;
const int N = 200010;
const int offset = 100002;

long long tree[N];
long long threshold;
int in[offset + 5];
int n;

long long query(int k) {
    long long s = 0;
    while(k >= 1) {
        s += tree[k];
        k -= k&-k;
    }
    return s;
}

void add(int k, int v) {
    while(k < N) {
        tree[k] += v;
        k += k&-k;
    }
}


bool f(int x) {
    memset(tree, 0, sizeof tree);
    long long cnt = 0;
    int sum = offset;
    add(offset, 1);
    for(int i = 1; i <= n; i++) {
        if(in[i] >= x) sum++;
        else sum--;
        cnt += query(sum);
        add(sum, 1);
    } 

    return cnt >= threshold;
}

int main() {
    scanf("%d", &n);
    int maxa = 0;
    for(int i = 1; i <= n; ++i) {
        scanf("%d", in + i);
        maxa = max(in[i], maxa);
    }

    threshold = n * 1LL * (n + 1) / 2LL;
    threshold = (threshold + 1LL) / 2LL;

    int ans = 0;
    for(int b = maxa; b >= 1; b /= 2) {
        while(ans + b <= maxa and f(ans + b)) ans += b;
    }

    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"
using namespace std;
const int N = 200010;
const int offset = 100002;

long long tree[N];
long long threshold;
int in[offset + 5];
int n;

long long query(int k) {
    long long s = 0;
    while(k >= 1) {
        s += tree[k];
        k -= k&-k;
    }
    return s;
}

void add(int k, int v) {
    while(k < N) {
        tree[k] += v;
        k += k&-k;
    }
}


bool f(int x) {
    memset(tree, 0, sizeof tree);
    long long cnt = 0;
    int sum = offset;
    add(offset, 1);
    for(int i = 1; i <= n; i++) {
        if(in[i] >= x) sum++;
        else sum--;
        cnt += query(sum);
        add(sum, 1);
    } 

    return cnt >= threshold;
}

int main() {
    scanf("%d", &n);
    int maxa = 0;
    for(int i = 1; i <= n; ++i) {
        scanf("%d", in + i);
        maxa = max(in[i], maxa);
    }

    threshold = n * 1LL * (n + 1) / 2LL;
    threshold = (threshold + 1LL) / 2LL;

    int ans = 0;
    for(int b = maxa; b >= 1; b /= 2) {
        while(ans + b <= maxa and f(ans + b)) ans += b;
    }

    printf("%d\n", ans);
}
```