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 collections import defaultdict as dd
t=1
d=dd(list)
for _ in range(t):
    n,k=map(int,input().split())
    arr=list(map(int,input().split()))
    arr.sort()
    for i in range(n):
        j=arr[i]
        d[j]+=[0]
        co=1
        while(j!=0):
            j=j//2
            if(j in d):
                d[j]+=[d[j][-1]+co]
            else:
                d[j]+=[co]
            co+=1
    mini=999999999999
    #print(d)
    for i in d:
        if(len(d[i])>=k):
            mini=min(d[i][k-1],mini)
    print(mini)