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:
import math as mt
import sys,string
input=sys.stdin.readline
from collections import defaultdict
L=lambda : list(map(int,input().split()))
Ls=lambda : list(input().split())
M=lambda : map(int,input().split())
I=lambda : int(input())


n,k=M()
l=L()
x=set()
for i in l:
    r=i
    while(r):
        x.add(r)
        r=r//2
        x.add(r)
y=defaultdict(list)
ans1=10**10

for j in range(len(l)):
    ans=0
    d=l[j]
    while(d):
        y[d].append(ans)
        d//=2
        ans+=1
    if(d==0):
        y[d].append(ans)
w=list(y.keys())

ans=10**10
for i in range(len(w)):
    if(len(y[w[i]])>=k):
        ans=min(ans,sum(sorted(y[w[i]])[:k]))
print(ans)