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
import sys
range = xrange
input = raw_input


n, k = [int(x) for x in input().split()]
A = sorted(int(x) for x in input().split())

bigsum = sum(A)
bigcount = n

smallsum = 0
smallcount = 0

cur = 0
best = 10**18

i = 0
for a in set(A):    
    while i < n and A[i] == a:
        bigsum -= a
        bigcount -= 1
        cur += 1
        i += 1
    if cur >= k:
        best = 0
        break
    
    smallcost = (a - 1) * smallcount - smallsum
    bigcost = bigsum - (a + 1) * bigcount
    
    best = min(best, smallcost + bigcost + k - cur)

    if smallcount + cur >= k:
        best = min(best, smallcost + k - cur)

    if bigcount + cur >= k:
        best = min(best, bigcost + k - cur)

    smallsum += cur * a
    smallcount += cur
    cur = 0

print best
