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
from itertools import combinations
from math import ceil
N = int(input())
A = list(map(int, input().split()))
M = ceil(N*(N+1)/4)

class FenwickTree(object):
    def __init__(self, n):
        self._bit = [0] * (n+1)
        self.n = n+1
    
    def add(self, i, a):
        i += 1
        while i < self.n:
            self._bit[i] += a
            i += i & -i
    
    def sum(self, i):
        i += 1
        ret = 0
        while i > 0:
            ret += self._bit[i]
            i -= i & -i
        return ret

lo, hi = 0, max(A)+1
while lo < hi-1:
    mid = (lo+hi)//2
    B = [1 if a >= mid else -1 for a in A]
    C = []
    C.append(0)
    for b in B:
        C.append(C[-1]+b)
    cmin = min(C)
    for i in range(N+1):
        C[i] -= cmin  # offset
    cmax = max(C)+1
    fw = FenwickTree(cmax)
    cnt = 0
    for i in range(N+1):
        cnt += fw.sum(C[i])
        fw.add(C[i], 1)

    if cnt >= M:
        lo = mid
    else:
        hi = mid

print(lo)
