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:
import math
n,k = map(int,input().split())
l = list(map(int,input().split()))
l.sort()
d = {}
c = {}
for i in l:
    cnt = 0
    x = i
    while x != 0:
        if x in d:
            d[x] += 1

        else:
            d[x] = 1

        if x in c:
            if d[x] <= k:
                c[x] += cnt

        else:
            c[x] = cnt

        x = x//2
        cnt += 1

    if x in d:
        d[x] += 1

    else:
        d[x] = 1

    if x in c:
        if d[x] <= k:
            c[x] += cnt

    else:
        c[x] = cnt

ans = 10**18
# print(d)
for i in d:
    if d[i] >= k:
        ans = min(ans,c[i])

print(ans)