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
def mergecount(A):
    cnt = 0
    n = len(A)
    if n>1:
        A1 = A[:n>>1]
        A2 = A[n>>1:]
        cnt += mergecount(A1)
        cnt += mergecount(A2)
        i1 = 0
        i2 = 0
        for i in range(n):
            if i2 == len(A2):
                A[i] = A1[i1]
                i1 += 1
            elif i1 == len(A1):
                A[i] = A2[i2]
                i2 += 1
            elif A1[i1] <= A2[i2]:
                A[i] = A1[i1]
                i1 += 1
            else:
                A[i] = A2[i2]
                i2 += 1
                cnt += n//2 - i1
    return cnt
 
 
n = int(input())
l = [int(_) for _ in input().split()]
p = (n*n+n)/2
q = 0
if p%2 == 0:
    q = p//2
else:
    q = p//2 + 1
l_set = list(set(l))
l_sorted = sorted(l_set)
ans = 0
l_t = [0 for _ in range(n)]
s = [0 for _ in range(n+1)]
 
low = 0
high = len(l_sorted) - 1
 
 
while low <= high:
    mid = (low + high) // 2
    m = l_sorted[mid]
    for j in range(n):
        if l[j] < m:
            l_t[j] = -1
        else:
            l_t[j] = 1
    for k in range(n):
        s[k+1] = s[k] + l_t[k]
 
    x = mergecount(s)
    if p-x >= q:
        low = mid+1
        ans = max(m,ans)
    else:
        high = mid-1
print(ans)