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 = list(map(int, input().split()))

# n = 4
# A = [i+1 for i in range(n)]

class Bit:
    def __init__(self, n):
        self.size = n
        self.tree = [0] * (n + 1)
  
    def sum(self, i):
        s = 0
        while i > 0:
            s += self.tree[i]
            i -= i & -i
        return s
  
    def add(self, i, x):
        while i <= self.size:
            self.tree[i] += x
            i += i & -i

#print((n*(n+1))//(2*2))

le = -1
ri = max(A) + 1
mid = (le + ri) // 2
while ri - le > 1:
    S = [0] * n
    S[0] = -1 + 2 * int(A[0] >= mid)
    for i in range(n-1):
        S[i+1] = S[i] - 1 + 2 * int(A[i+1] >= mid)
    bit = Bit(n+2)
    ans = 0
    for i in range(n):
        if S[i] >= 0:
            ans += 1

    #print("S OK")

    Smin = min(S)
    S = [S[i] - Smin + 1 for i in range(n)]

    #print("S OK2")
    #print(S)
    for i, p in enumerate(S):
        
        bit.add(p, 1)

        ans += bit.sum(p) - 1  # 「S[i] >= S[i]」をカウントしないように1を引く
        #print(i,ans)


    #print(le, ri, mid, ans)

    if ans >= (n*(n+1))//(2*2):
        le = mid
    else:
        ri = mid
    mid = (le + ri) // 2

print(le)
    
