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
N = int(input())
A = [int(x) for x in input().split()]
total_medians = N*(N+1)//2
# n件のデータ
# (上側中央値 >= x) iff (a > x となる a が(n-1)//2以下)

def BIT_update(BIT,x,value):
  # 番号xにvalueを加える
  L = len(BIT)
  while x < L:
    BIT[x] += value
    bottom_bit = x & (-x)
    x += bottom_bit
  return

def BIT_sum(BIT,x):
  # 番号1～xの和を求める
  result = 0
  while x:
    result += BIT[x]
    bottom_bit = x & (-x)
    x -= bottom_bit
  return result

def test(x):
  # 「答はx以下である」
  # 「xより真に大きな区間中央値が(total_medians-1)//2個以下」
  B = [0]
  for a in A:
    B.append(B[-1] + (1 if a>x else -1))
  m = min(B)-1
  B = [x-m for x in B]

  BIT = [0]*(N+1)
  
  cnt = 0
  for R in range(1,N+1):
    BIT_update(BIT,B[R-1],1)
    cnt += BIT_sum(BIT,B[R]) # 右端がRとなる(L,R)対
  return cnt <= (total_medians-1)//2

AA = sorted(A)
left = -1 # 答はAA[left]より大きい
right = len(AA)-1 # 答はAA[right]以下である
while right - left > 1:
  mid = (left+right)//2
  if test(AA[mid]):
    right = mid
  else:
    left = mid
answer = AA[right]

print(answer)
