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(i) for i in input().split()]
data = [int(i) for i in input().split()]

dic = [[0]*20 for j in range(200001)] 

for d in data:
    s = 0
    while d:
        dic[d][s] += 1
        d >>= 1
        s += 1
    dic[0][s] += 1

mn = 1<<30

for d in dic:
    if sum(d) >= k:
        left = k
        val = 0
        for i in range(20):
            if d[i] >= left:
                val += i * (left)
                break
            else:
                val += i * d[i]
                left -= d[i]

        if val < mn:
            mn = val
print(mn)