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 <bits/stdc++.h>
using namespace std;
using ll = long long;

template< typename T >
struct BinaryIndexedTree {
	vector< T > data;

	BinaryIndexedTree(int sz) {
		data.assign(++sz, 0);
	}

	T sum(int k) {
		T ret = 0;
		for(++k; k > 0; k -= k & -k) ret += data[k];
		return (ret);
	}

	void add(int k, T x) {
		for(++k; k < data.size(); k += k & -k) data[k] += x;
	}
};

signed main() {
	ll N;
	scanf("%lld", &N);
	vector<ll> a(N, 0);
	for(ll i=0; i<N; i++) scanf("%lld", &a[i]);
	if(N==1){
		printf("%lld\n", a[0]);
		return 0;
	}

	ll left = 0;
	ll right = 1e9+1;
	auto isOK = [&](ll val) -> bool {
		vector<ll> b(N,0);
		for(ll i=0; i<N; i++) b[i] = (a[i]<val ? -1 : 1);
		vector<ll> rui(N+1, 0);
		for(ll i=0; i<N; i++) rui[i+1] = rui[i] + b[i];
		for(ll i=0; i<N+1; i++) rui[i] += N;

		BinaryIndexedTree<ll> bit(2*N);
		ll tentosu=0;
		for(ll i=0; i<N+1; i++) {
			tentosu += i - bit.sum(rui[i]);
			bit.add(rui[i], 1);
		}
		ll t = N*(N+1)/2 - tentosu;
		return t >= N*(N+1)/2/2;
	};

	while(right - left > 1) {
		ll mid = left + (right - left) / 2;
		if(isOK(mid)) left = mid;
		else right = mid;
	}

	printf("%lld\n", left);
	return 0;
}
