#include <bits/stdc++.h>
#define mem(x,v) memset(x,v,sizeof(x)) 
#define go(i,a,b)  for (int i = a; i <= b; i++)
#define og(i,a,b)  for (int i = a; i >= b; i--)
#define MID(a,b) (a + b) >> 1
#define lson now << 1
#define rson now << 1 | 1
#define low(x) (x & (-x))
using namespace std;
typedef long long LL;
const double EPS = 1e-10;
const int INF = 1e9+100;
const int N = 1e6+10;
int a[N],b[N],c[N];
LL f[N];
int n,tt;
void Add(int x){
	while(x <= 5*n){
		f[x] += 1;
		x += low(x);
	}
}
int Query(int x){
	int ans = 0;
	while(x > 0) ans += f[x], x -= low(x);
	return ans; 
}
bool find(int k){
	mem(f,0);
	LL Ans = 0;	b[0] = 0;
	go(i,1,n) if (a[i] >= c[k]) b[i] = b[i-1] + 1; else b[i] = b[i-1];
	go(i,0,n) {
		b[i] = 2*b[i] - i + n + 1;
		Ans += Query(b[i]);
		Add(b[i]);
	}
	LL t = (1LL * n * (n + 1)) / 2;
	if (Ans >= (t+1)/2) return 1; 
	return 0;
}

int main(){
	scanf("%d",&n);
	int l = 1, r = n;
	go(i,1,n){
		scanf("%d",&a[i]);
		c[i] = a[i];
	}
	sort(c+1,c+n+1);
	int ans = 0;
	while(l <= r){
		int mid = MID(l,r);
		if (find(mid)) l = mid+1,ans = c[mid]; else r = mid - 1;
	}
	printf("%d\n",ans);
	return 0;
}
### Prompt

Please create a solution in CPP 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>
#define mem(x,v) memset(x,v,sizeof(x)) 
#define go(i,a,b)  for (int i = a; i <= b; i++)
#define og(i,a,b)  for (int i = a; i >= b; i--)
#define MID(a,b) (a + b) >> 1
#define lson now << 1
#define rson now << 1 | 1
#define low(x) (x & (-x))
using namespace std;
typedef long long LL;
const double EPS = 1e-10;
const int INF = 1e9+100;
const int N = 1e6+10;
int a[N],b[N],c[N];
LL f[N];
int n,tt;
void Add(int x){
	while(x <= 5*n){
		f[x] += 1;
		x += low(x);
	}
}
int Query(int x){
	int ans = 0;
	while(x > 0) ans += f[x], x -= low(x);
	return ans; 
}
bool find(int k){
	mem(f,0);
	LL Ans = 0;	b[0] = 0;
	go(i,1,n) if (a[i] >= c[k]) b[i] = b[i-1] + 1; else b[i] = b[i-1];
	go(i,0,n) {
		b[i] = 2*b[i] - i + n + 1;
		Ans += Query(b[i]);
		Add(b[i]);
	}
	LL t = (1LL * n * (n + 1)) / 2;
	if (Ans >= (t+1)/2) return 1; 
	return 0;
}

int main(){
	scanf("%d",&n);
	int l = 1, r = n;
	go(i,1,n){
		scanf("%d",&a[i]);
		c[i] = a[i];
	}
	sort(c+1,c+n+1);
	int ans = 0;
	while(l <= r){
		int mid = MID(l,r);
		if (find(mid)) l = mid+1,ans = c[mid]; else r = mid - 1;
	}
	printf("%d\n",ans);
	return 0;
}
```