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:
tmp = input().split()
n= int(tmp[0])
k = int(tmp[1])
tmp = input().split()
tree = [ list() for i in range(200001) ]
minlength = 2000001
for value in tmp:
    value = int(value)
    length = 0
    while(value != 0):
        if len(tree[value]) < k:
            tree[value].append(length)
            if len(tree[value]) == k and minlength > sum(tree[value]):
                minlength = sum(tree[value])
        else:
            if max(tree[value]) > length:
                tree[value].remove(max(tree[value]))
                tree[value].append(length)
                if minlength > sum(tree[value]):
                    minlength = sum(tree[value])
        value = value // 2
        length += 1
print(minlength)