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 heapq import *
N, K = map(int,input().split())
heap = []
for i in list(map(int,input().split())):
    heappush(heap, (-i, 0))

cnt = 1
div_cnt = 0
prv = -1
ans = 10**10
while heap:
    num, div_c = heappop(heap)
    num = -num
    if prv == num:
        cnt += 1
        div_cnt += div_c
    else:
        prv = num
        div_cnt = div_c
        cnt = 1
    if cnt == K:
        ans = min(ans, div_cnt)
    if num != 0:
        heappush(heap, ( - (num//2), div_c + 1)  )
print(ans)