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
#include <cstdio>
#include <iostream>
#include <vector>
using namespace std;

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

long long merge_sort(int l,int r){
	if (l == r) return 0;
	int mid = (l+r)/2;
	long long ret = merge_sort(l,mid) + merge_sort(mid+1,r);
	int i = l;
	int j = mid+1;
	int p = l;
	while (i <= mid || j <= r) {
		if (j > r || (i <= mid && s[i] <= s[j])){
			b[p++] = s[i++];
		}else{
			b[p++] = s[j++];
			ret += mid-i+1;
		}
	}
	for (int i = l; i <= r; i++) s[i] = b[i];
	return ret;
}

bool check(int x) {
	s[0] = 0;
	for (int i = 1; i <= n; i++) s[i] = s[i-1] + (a[i] < x? -1 : 1);
	return merge_sort(0,n)*2 <= (long long) n*(n+1)/2;
}

int main(){
	scanf("%d",&n);
	vector<int> vec;
	for (int i = 1; i <= n; i++) {
		scanf("%d",&a[i]);
		vec.push_back(a[i]);
	}
	sort(vec.begin(), vec.end());
	/*for (int i = 0; i < vec.size(); i++) {
		printf("%d ",vec[i]);
	}
	printf("\n");*/
	vec.erase(unique(vec.begin(),vec.end()),vec.end());
	int l = 0, r = vec.size() - 1;
	while (l < r) {
		int mid = (l+r+1)/2;
		if (check(vec[mid])) l = mid;
		else r = mid -1;
	}
	printf ("%d\n",vec[l]);
	return 0;
}