#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#define itn int
#define reaD read
typedef long long ll;
#define N 100005
using namespace std;

ll n, a[N], l, r, t[N << 1], sum[N]; 

inline int read()
{
	int x = 0, w = 1; char c = getchar();
	while(c < '0' || c > '9') { if (c == '-') w = -1; c = getchar(); }
	while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
	return x * w;
}

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

void add(int x, long long y) { for(int i = x; i <= 200000; i += lowbit(i)) t[i] += y; }

long long query(int x) { long long res = 0; for(int i = x; i; i -= lowbit(i)) res += t[i]; return res; }

bool check(long long mid)
{
	memset(t, 0, sizeof(t));
	long long cnt = 0;
	for(int i = 1; i <= n; i++) sum[i] = sum[i - 1] + (a[i] >= mid ? 1 : -1);
	for(int i = 0; i <= n; i++)
	{
		cnt += query(sum[i] + 100000);
		add(sum[i] + 100000, 1); 
	}
	return cnt >= 1ll * n * (n + 1) / 4; 
}

int main()
{
	n = read();
	for(int i = 1; i <= n; i++) r = max(r, a[i] = read());
	while(l <= r)
	{
		int mid = (l + r) >> 1;
		if(check(mid)) l = mid + 1;
		else r = mid - 1; 
	}
	printf("%lld\n", r); 
	return 0;
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <vector>
#define itn int
#define reaD read
typedef long long ll;
#define N 100005
using namespace std;

ll n, a[N], l, r, t[N << 1], sum[N]; 

inline int read()
{
	int x = 0, w = 1; char c = getchar();
	while(c < '0' || c > '9') { if (c == '-') w = -1; c = getchar(); }
	while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
	return x * w;
}

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

void add(int x, long long y) { for(int i = x; i <= 200000; i += lowbit(i)) t[i] += y; }

long long query(int x) { long long res = 0; for(int i = x; i; i -= lowbit(i)) res += t[i]; return res; }

bool check(long long mid)
{
	memset(t, 0, sizeof(t));
	long long cnt = 0;
	for(int i = 1; i <= n; i++) sum[i] = sum[i - 1] + (a[i] >= mid ? 1 : -1);
	for(int i = 0; i <= n; i++)
	{
		cnt += query(sum[i] + 100000);
		add(sum[i] + 100000, 1); 
	}
	return cnt >= 1ll * n * (n + 1) / 4; 
}

int main()
{
	n = read();
	for(int i = 1; i <= n; i++) r = max(r, a[i] = read());
	while(l <= r)
	{
		int mid = (l + r) >> 1;
		if(check(mid)) l = mid + 1;
		else r = mid - 1; 
	}
	printf("%lld\n", r); 
	return 0;
}
```