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

a = [int(ai) for ai in input().split()]

possibilities = []
for ai in a:
    moves = 0
    while ai != 0:
        possibilities.append((ai,  moves))
        ai = ai // 2
        moves += 1
    possibilities.append((ai, moves))
possibilities.sort()

i = 0
minOps = inf
while i < len(possibilities):
    same = possibilities[i][0]
    sameElems = 0
    operations = 0
    j = i
    while(j < len(possibilities) and possibilities[j][0] == same):
        sameElems += 1
        operations += possibilities[j][1]
        if sameElems == k:
            break
        j += 1
    if sameElems == k and operations < minOps:
        minOps = operations
    j = i
    while(j < len(possibilities) and possibilities[j][0] == same):
        j += 1
    i = j
print(minOps)
