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 deque
import collections
import sys

def inp():
    return sys.stdin.readline().strip()
n,k=map(int,inp().split())
a=list(map(int,inp().split()))
d=collections.Counter(a)
b=list(d.keys())
mn=float('inf')
b.sort()
for i in range(max(b)+1):
    val=d.get(i,0)
    ans=0
    for j in b:
        if j>i:
            ct=0
            temp=j
            while temp>i:
                temp//=2
                ct+=1 
            if temp==i:
                if val<k and d[j]+val>=k:
                    req=k-val
                    val=k
                    ans+=req*ct
                    break
                else:
                    val+=d[j]
                    ans+=d[j]*ct
                
    if val>=k:
        mn=min(ans,mn)
print(mn)