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 sys import stdin,stdout
from math import gcd,sqrt,inf
from collections import deque
input=stdin.readline
R=lambda:map(int,input().split())
I=lambda:int(input())
S=lambda:input().rstrip('\n')
L=lambda:list(R())
P=lambda x:stdout.write(x)
hg=lambda x,y:((y+x-1)//x)*x
pw=lambda x:1 if x==1 else 1+pw(x//2)
chk=lambda x:chk(x//2) if not x%2 else True if x==1 else False
N=inf
n,k=R()
a=[deque() for i in range(2*10**5+7)]
v=[0]*(2*10**5+7)
sm=[0]*(2*10**5+7)
for i in sorted(R(),reverse=True):
	a[i].appendleft(0)
	v[i]+=1
	if v[i]>k:
		v[i]=k
		sm[i]-=a[i].pop()
	stp=0
	while i:
		i//=2
		stp+=1
		a[i].appendleft(stp)
		v[i]+=1
		sm[i]+=stp
		if v[i]>k:
			sm[i]-=a[i].pop()
			v[i]=k
ans=inf
for i in range(2*10**5+7):
	if v[i]==k:ans=min(sm[i],ans)
print(ans)