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,m=list(map(int,input().split()))
a=list(map(int,input().split()))
d=defaultdict(lambda:[])
for i in a:
    s=0
    while(i!=0):
        d[i]=d[i]+[s]
        s=s+1
        i=i//2
        if(i==0):
            d[0]=d[0]+[s]
#print(d)
k=list(d.keys())
for i in range(0,len(k)):
    d[k[i]]=sorted(list(d[k[i]]))
mn=9999999999999999
for i in range(0,len(k)):
    if(len(d[k[i]])>=m):
        mn=min(mn,sum(d[k[i]][0:m]))
print(mn)