There are *n* kangaroos with pockets. Each kangaroo has a size (integer number). A kangaroo can go into another kangaroo's pocket if and only if the size of kangaroo who hold the kangaroo is at least twice as large as the size of kangaroo who is held.

Each kangaroo can hold at most one kangaroo, and the kangaroo who is held by another kangaroo cannot hold any kangaroos.

The kangaroo who is held by another kangaroo cannot be visible from outside. Please, find a plan of holding kangaroos with the minimal number of kangaroos who is visible.
The first line contains a single integer — *n* (1 ≤ *n* ≤ 5·10<sup class="upper-index">5</sup>). Each of the next *n* lines contains an integer *s*<sub class="lower-index">*i*</sub> — the size of the *i*-th kangaroo (1 ≤ *s*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">5</sup>).

sizes = []
n = int(input())
for i in range(n):
    sizes.append(int(input()))
    sizes.sort()
    
    
    
# once array is sorted now kangroos  whose size is sizes[i] can only hold  kangroo whose sizes[j]
# where j < i and sizes[i]>=2*sizes[j]
# greedy move  first try to hide kangroos with larger sizes 
# greedy move hide the largest kangroo possible 
#  greedy move try to fill the pocket of the largest kangroo first with the largest kangroo it can fit 
# think about for each kangroo find the number of kangroos it can fit inside its pocket 
# also we could have focused on how many kangroos can fit a kangroo 
def min_visible(n,sizes):
    # binary serach n is always true 
    # 0 always false
    # it  is monotonic cannot have fewer than x kangroos can always have more 
 
        
        # do some calculation 
        # check if possible to have only mid no. of kangroos visible 
        # if yes change l 
        # if no change r 
        
        
    # ASSUME that the last half are never  hidden and they are the ones that hid the smaller ones but what if there are odd number of kangroos
    # if 2n+1 kagroos
    # maximum we can hide n only 
    # we leave the first n+1 from the starting and only the last n can hide we can safely say 
    # two pointer approach the first one from strating second one from the middle 
    # each iteration we increase the pointer from starting by one and also the pointer from middle 
    # if n is 2k then 0 to k-1 leave so start from k and go till end 
    # if n = 2k+1 then leave the 0 ....k and start from k+1 go till end 
    # k = n//2 
    visible = n
    if n%2 ==0:
        start = n//2
    else:
        start = (n//2)+1
    for i in range(n):
        
        while(start!=n-1):
            if sizes[start]>=2*sizes[i]:
                visible-=1
                break
            else:
                start+=1
        if start ==n-1:
            break
        # why i inrange(n) doubt 
    return visible
            
            
#        for j in range(start,n):
            
   
        
        
        
        
        
        
print(min_visible(n,sizes))
    
    
    
    
    
    
    
    
    