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:
def inti_dic(d,no,opertions):
 if no in d:
  a=d[no]
  if a[0]<k:#if  no less than desire k than only add opeartion
   a[0]+=1#otherwise operation would be increase
   a[1]+=op
   d[no]=a
 else:d[no]=[1,opertions]

n,k=map(int,input().split())
arr=list(map(int,input().split()))
d={}
arr=sorted(arr)
for i in arr:
 no,op=i,0
 inti_dic(d,no,op)#total opertion to make this no
 while no!=0:
   no=no//2
   op+=1
   inti_dic(d,no,op)#total opeartion to make this no
minn=10**15
for i in d:
  a=d[i]
  if a[0]>=k:minn=min(a[1],minn)
print(minn)