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:
import collections
n,k=map(int,input().split())
A=[int(i) for i in input().split()]
ans=float("inf")
D1,D2=collections.defaultdict(list),collections.defaultdict(int)
for i in range(n):
  a=A[i]
  c=0
  D1[a].append(c)
  D2[a]+=1
  while a>0:
    a//=2
    c+=1
    D1[a].append(c)
    D2[a]+=1
for i in range(max(A)+1):
  if D2[i]>=k:
    D1[i].sort()
    ans=min(ans,sum(D1[i][:k]))
print(ans)