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 PYTHON):

Solution:
import sys
n,k=map(int,raw_input().split())
l=list(map(int,raw_input().split()))
l.sort()

d={}
for i in xrange(0,2*(10**5)+1):
    key=i
    d[key]=[0,[]]

for i in xrange(len(l)):
    key=l[i]
    if key in d:
        pass
    else:
        d[key]=[1,[]]

for i in d:
    if d[i][0]>=k:
        print 0
        sys.exit()

for i in xrange(len(l)):
    x=l[i]
    c=0
    d[x][0]+=1
    d[x][1].append(c)
    while x!=0:
        x/=2
        c+=1
        d[x][0]+=1
        d[x][1].append(c)
        

ans=10**9
for i in d:
    if len(d[i][1])>=k:
        s=sum(d[i][1][:k])
        ans=min(ans,s)
print ans