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

const int MAX = 200000;
const int ADD = 100001;

int a[100010];
int T[200010];
int n;

void add(int x, int v){
	while (x <= MAX){
		T[x] += v;
		x += x & -x;
	}
}

int query(int x){
	int res = 0;
	while (x){
		res += T[x];
		x -= x & -x;
	}
	return res;
}

int sum[100010];

int check(int x){
	memset(T, 0, sizeof(T));
	long long res = 0;
	add(ADD, 1);
	for (int i = 0; i < n; i++){
		sum[i] = a[i] >= x ? 1 : -1;
		if (i) sum[i] += sum[i - 1];
//		cout << "  sum:" << sum[i] << endl;
		res += query(sum[i] + ADD);
//		cout << "  " << i << " " << query(sum[i] + ADD) << endl;
		add(sum[i] + ADD, 1);
	}
//	cout << x << " " << res << endl;
	return res >= 1ll * n * (n + 1) / 4;
}

int main(){
	scanf("%d", &n);
	int maxj = 0;
	for (int i = 0; i < n; i++){
		scanf("%d", &a[i]);
		maxj = max(maxj, a[i]);
	}
	int l = 0, r = maxj;
//	cout << check(8) << endl;
	while (l < r){
		int mid = (l + r + 1) >> 1;
//		cout << l << " " << r << " " << mid << " " << check(mid) << endl;
		if (check(mid)) l = mid;
		else r = mid - 1;
	}
	printf("%d\n", l);
	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
#include <bits/stdc++.h>
using namespace std;

const int MAX = 200000;
const int ADD = 100001;

int a[100010];
int T[200010];
int n;

void add(int x, int v){
	while (x <= MAX){
		T[x] += v;
		x += x & -x;
	}
}

int query(int x){
	int res = 0;
	while (x){
		res += T[x];
		x -= x & -x;
	}
	return res;
}

int sum[100010];

int check(int x){
	memset(T, 0, sizeof(T));
	long long res = 0;
	add(ADD, 1);
	for (int i = 0; i < n; i++){
		sum[i] = a[i] >= x ? 1 : -1;
		if (i) sum[i] += sum[i - 1];
//		cout << "  sum:" << sum[i] << endl;
		res += query(sum[i] + ADD);
//		cout << "  " << i << " " << query(sum[i] + ADD) << endl;
		add(sum[i] + ADD, 1);
	}
//	cout << x << " " << res << endl;
	return res >= 1ll * n * (n + 1) / 4;
}

int main(){
	scanf("%d", &n);
	int maxj = 0;
	for (int i = 0; i < n; i++){
		scanf("%d", &a[i]);
		maxj = max(maxj, a[i]);
	}
	int l = 0, r = maxj;
//	cout << check(8) << endl;
	while (l < r){
		int mid = (l + r + 1) >> 1;
//		cout << l << " " << r << " " << mid << " " << check(mid) << endl;
		if (check(mid)) l = mid;
		else r = mid - 1;
	}
	printf("%d\n", l);
	return 0;
}

```