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
# coding: utf-8
# Your code here!

import sys
sys.setrecursionlimit(10**6)
readline = sys.stdin.readline 

n = int(input())
a = [int(i) for i in readline().split()]

def binarysearch(ok,ng,isOK): #二分探索
    while(abs(ok-ng) > 1):
        mid = (ok + ng)//2
        if isOK(mid): ok = mid
        else: ng = mid
    return ok

from collections import Counter
def isOK(x): #中央値はx以上？
    c = [1 if i >= x else -1 for i in a]
    res = Counter([0])
    now = 0
    less = 1
    ans = 0
    for i in range(n):
        if c[i] > 0:
            less += res[now+1]
        else:
            less -= res[now]
        now += c[i]
        ans += less
        less += 1
        res[now] += 1
    return ans >= (n*(n+1)//2+1)//2      


ans = binarysearch(0,10**9+1,isOK)
print(ans)

