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
n,k=map(int,input().split())
ar=[int(x) for x in input().split()]
ar.sort()
d={}
for i in ar:
    if(i not in d):
        d[i]=1
    else:
        d[i]+=1
bl=False
d=defaultdict(list)
for i in ar:
    count=0
    x=i
    while(x):
        d[x].append(count)
        x//=2
        count+=1

ans=1000000000
for i in d.keys():
    d[i].sort()
    for j in range(len(d[i])):
        if(j>0):
            d[i][j]+=d[i][j-1]
    if(len(d[i])>=k):
        ans=min(ans,d[i][k-1])
print(ans)