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:
"""
NTC here
"""
from sys import setcheckinterval,stdin
setcheckinterval(1000)

#print("Case #{}: {} {}".format(i, n + m, n * m))

iin=lambda :int(stdin.readline())
lin=lambda :list(map(int,stdin.readline().split()))
from collections import defaultdict
n,k=lin()
a=lin()
a.sort()
sol=[[0,0] for i in range(2*10**5+1)]
for i in a:
    ch=0
    x=i
    if sol[i][0]<k:
        sol[i][0]+=1
    while x>0:
        x//=2
        ch+=1
        if sol[x][0]<k:
            sol[x][0]+=1
            sol[x][1]+=ch 
ans=99999999999
#print(sol[:10])
for i,j in sol:
    if i>=k:
        ans=min(ans,j)
print(ans)