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:
#582_D1

import math

l = [int(i) for i in input().split(" ")]
n = l[0]
k = l[1]

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

ln = sorted(ln)

m = 1000000000

def findPart(n, a, d):
    s1 = []
    s2 = []
    s3 = []
    for i in range(0, len(a)):
        n = math.floor(math.log(a[i], 2)) + 1
        if n - d < 0:
            s3.append(a[i])
            continue
        bt = a[i] & (1 << (n - d))
        if bt == 0:
            s1.append(a[i])
        else:
            s2.append(a[i])
    ops = []
    for i in range(0, len(a)):
        n = math.floor(math.log(a[i], 2)) + 1
        ops.append(n - (d - 1))
    m = sum(ops[:k])
    if len(a) - len(s3) >= k:
        if len(s1) >= k and len(s2) >= k:
            m = min(m, findPart(n * 2, s1, d + 1), findPart(n * 2 + 1, s2, d + 1))
        elif len(s1) >= k:
            m = min(m, findPart(n * 2, s1, d + 1))
        elif len(s2) >= k:
            m = min(m, findPart(n * 2 + 1, s2, d + 1))
    return m
print(findPart(1, ln, 2))