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, m = map(int, input().split())
a = list(map(int, input().split()))
a.sort()
b = {}
c = {}
mm = 10e10
for i in range(n):
    k = a[i]
    t = 0
    while k > 0:
        if k not in b or b[k] < m:
            b[k] = b.get(k, 0) + 1
            c[k] = c.get(k, 0) + t
            if b[k] == m:
                mm = min(mm, c[k])
        t += 1
        k //= 2
    else:
        if 0 not in b or b[0] < m:
            b[0] = b.get(0, 0) + 1
            c[0] = c.get(0, 0) + t
        if b[0] == m:
            mm = min(mm, c[0])

print(mm)