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
# -*- coding: utf-8 -*-

import sys
import copy

sys.setrecursionlimit(1000000)

# input = sys.stdin.readline
# from collections import defaultdict
# import bisect

# ~~~~~~~~~~~~~~~~~~~~~_(＾～＾ ｣ ∠)_~~~~~~~~~~~~~~~~~~~~~


n, k = map(int, input().split())

a = list(map(int, input().split()))

B = [[] for i in range(200200)]

for i in range(n):
    cnt = 0
    while a[i] > 0:
        B[a[i]].append(cnt)
        cnt += 1
        a[i] //= 2
    B[a[i]].append(cnt)

ans = 1e20
for i in range(len(B)):
    if len(B[i]) >= k:
        B[i].sort()
        ans = min(ans, sum(B[i][0:k]))

print(ans)
