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:
# TAIWAN NUMBER ONE!!!!!!!!!!!!!!!!!!!
# TAIWAN NUMBER ONE!!!!!!!!!!!!!!!!!!!
# TAIWAN NUMBER ONE!!!!!!!!!!!!!!!!!!!
from sys import stdin, stdout
from collections import defaultdict
from collections import deque
import math
import copy
 
#T = int(input())
#N = int(input())
#s1 = input()
#s2 = input()
N,K = [int(x) for x in stdin.readline().split()]
arr = [int(x) for x in stdin.readline().split()]
arr.sort()

freq = {}

for i in range(N):
    num = arr[i]
    if num not in freq:
        freq[num] = []
        
    round = 0
    freq[num].append(0)
    while num!=0:
        round += 1
        num = num//2
        if num not in freq:
            freq[num] = []
            
        freq[num].append(round)

res = 999999999999
for key in freq:
    if len(freq[key])<K:
        continue
    else:
        s = sum(freq[key][:K])
        res = min(res,s)
        
print(res)