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
import io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import sys
n,k=list(map(int,input().split()))
arr=list(map(int,input().split()))
arr.sort()
a=arr[-1]
count=[0]*(a)
for i in range(n):
    count[arr[i]-1]+=1
if max(count)>=k:
    print(0)
else:
    matrix=[[] for i in range(n)]
    steps=[[] for i in range(a)]
    for i in range(n):
        temp=arr[i]
        s=0
        while temp>0:
            matrix[i].append(temp)
            steps[temp-1].append(s)
            if temp!=arr[i]:
                count[temp-1]+=1
            temp=temp//2
            s+=1
    lis=[]
    l=0
    for i in range(a):
        if count[i]>=k:
            lis.append(i)
            l+=1
    ans=sys.maxsize
    for i in range(l):
        steps[lis[i]].sort()
        ans=min(ans,sum(steps[lis[i]][:k]))
    print(ans)
        