Problem: You are given the array a consisting of n elements and the integer k ≤ n.

You want to obtain at least k equal elements in the array a. In one move, you can make one of the following two operations:

  * Take one of the minimum elements of the array and increase its value by one (more formally, if the minimum value of a is mn then you choose such index i that a_i = mn and set a_i := a_i + 1); 
  * take one of the maximum elements of the array and decrease its value by one (more formally, if the maximum value of a is mx then you choose such index i that a_i = mx and set a_i := a_i - 1). 



Your task is to calculate the minimum number of moves required to obtain at least k equal elements in the array.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a and the required number of equal elements.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

Output

Print one integer — the minimum number of moves required to obtain at least k equal elements in the array.

Examples

Input


6 5
1 2 2 4 2 3


Output


3


Input


7 5
3 3 2 1 1 1 3


Output


4

Solution (in PYTHON3):

Solution:



from math import log2
from collections import defaultdict

n,k = map(int,input().split())


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

l.sort()
z = l[-1]//2
hash = defaultdict(list)
seti = defaultdict(set)

for i in range(l[-1]+1):
    z = i
    while True:
        seti[i].add(z)
        if z == 0:
            break
        z = z//2

z = l[-1]
for i in range(z+1):

    for j in range(len(l)):
        if l[j]>=i:
            temp = i
            if i == 0:
                i+=1
            ka = log2(l[j]//i)


            if temp in seti[l[j]]:
              if temp == 0:
               ka+=1
               i-=1
              hash[i].append(int(ka))


mini = 10**18
for i in hash:
    hash[i].sort()
    if len(hash[i])>=k:

      mini = min(sum(hash[i][:k]),mini)

print(mini)