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:
n, k = [int(s) for s in input().split(" ")]
A = [int(s) for s in input().split(" ")]
divs = [0]*len(A)
maximum = max(A)
ans = float('inf')
while maximum > 0:
    if A.count(max(A)) >= k:
        D = [divs[i] for i in range(len(A)) if A[i] == maximum]
        D.sort()
        ans = min (ans, sum(D[:k]))
    max_index = A.index(max(A))
    A[max_index] = int(A[max_index] / 2 )
    maximum = max(A)
    divs[max_index] += 1
D = [divs[i] for i in range(len(A)) if A[i] == maximum]
D.sort()
ans = min (ans, sum(D[:k]))
print(ans)