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

cands = list(sorted(set(src)))

def enough(i):
    bit = [0] * (2*N+10)
    def bit_add(a,w):
        x = a
        while x <= (2*N+9):
            bit[x] += w
            x += (x & -x)
    def bit_sum(a):
        x = a
        ret = 0
        while x > 0:
            ret += bit[x]
            x -= (x & -x)
        return ret

    cums = [0]
    for a in src:
        tmp = 1 if a >= cands[i] else -1
        cums.append(cums[-1] + tmp)

    cnt = 0
    for i in range(N+1):
        a = cums[i] + N+5
        cnt += bit_sum(a)
        bit_add(a, 1)
    return cnt*2 >= N*(N+1)//2

ok = 0
ng = len(cands)
while ng - ok > 1:
    m = (ok+ng)//2
    if enough(m):
        ok = m
    else:
        ng = m
print(cands[ok])