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;

const int MAXN = 1.1e5;
int N;
int A[MAXN];

int bit[MAXN*2];
void reset() {
	for (int i = 0; i <= 2*N; i++) {
		bit[i] = 0;
	}
}
int query(int a) {
	assert(a <= 2 * N);
	int res = 0;
	for (; a; a -= a & (-a)) {
		res += bit[a];
	}
	return res;
}
void update(int a) {
	assert(a <= 2 * N);
	a++;
	for (; a <= 2 * N; a += a & (-a)) {
		bit[a] ++;
	}
}

ll go(ll v) {
	// we want to count # sequences where median is < v
	// # sequences where (#<v) is > (#>=v)
	ll res = 0;
	int pref = N;
	reset();
	update(pref);
	for (int i = 0; i < N; i++) {
		if (A[i] < v) {
			pref ++;
		} else {
			pref --;
		}
		assert(pref >= 0);
		res += query(pref);
		update(pref);
	}
	return res;
}

int M;
int V[MAXN];
int main() {
	ios_base::sync_with_stdio(0);
	cin >> N;
	for (int i = 0; i < N; i++) {
		cin >> A[i];
	}
	ll T = ll(N) * ll(N + 1) / 2;
	for (int i = 0; i < N; i++) {
		V[i] = A[i];
	}
	M = N;
	sort(V, V + M);
	M = int(unique(V, V + M) - V);
	int mi = 0, ma = M;
	while (ma - mi > 1) {
		int md = (mi + ma) / 2;
		if (go(V[md]) * 2 <= T) {
			mi = md;
		} else {
			ma = md;
		}
	}
	cout << V[mi] << '\n';

	return 0;
}
