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:
from math import inf
nk = input().split()
n = int(nk[0])
k = int(nk[1])

a = [int(ai) for ai in input().split()]
a.sort()
possibilities = {}
for ai in a:
    moves = 0
    while ai != 0:
        if ai not in possibilities:
            possibilities[ai] = (1, moves)
        else:
            curr = possibilities[ai]
            if curr[0] != k:
                possibilities[ai] = (curr[0] + 1, curr[1] + moves)
        ai = ai // 2
        moves += 1
minOps = inf
for i in possibilities:
    curr = possibilities[i]
    if curr[0] == k:
        if curr[1] < minOps:
            minOps = curr[1]

print(minOps)