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
import sys
MOD=pow(10,9)+7

def IN(f=0):
    if f==0:
        return ( [int(i) for i in sys.stdin.readline().split()] )
    else:
        return ( int(sys.stdin.readline()) )

n,k=IN()
a=IN()
b=[]

d=DD(list)
for x in a:
    i=0
    while(x!=0):
        d[x].append(i)
        x=x//2
        i+=1
    d[0].append(i)
    #print(d)

ans=9999999999999
for x in d:
    if len(d[x])>=k:
        r=d[x]
        r.sort()
        ans=min(ans,sum(r[:k]))
print(ans)