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 

n,k=map(int,input().split())
a=list(map(int,input().split()))
a.sort()

steps = [0]*200005 
count = [0]*200005 
ans = []
# 
#for i in range(n):
#    count[a[i]]+=1 

for i in range(n):
    x=a[i]
#    if count[x] == k :ans.append(steps[x])
#    print(math.floor(math.log2(a[i])+2))
    for j in range(math.floor(math.log2(a[i])+2)):
        if count[x] == k :
            ans.append(steps[x])

         
         
        count[x]+=1
        steps[x]+=j
        x=x//2
for i in range(200005):
    if count[i]==k:
        ans.append(steps[i])
        

print(min(ans))