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:
'''input
5 3
1 2 3 4 5
'''
from sys import stdin
from copy import deepcopy
from collections import deque, defaultdict


def get_arr(num):
	if num in freq:
		if len(freq[num]) >= k:
			return sum(freq[num][: k])
		else:
			return float('inf')
	else:
		return float('inf')
	



def get_min(arr):
	for i in freq:
		if freq[i] >= k:
			return 0
	return float('inf')


# main starts
n, k = list(map(int, stdin.readline().split()))
arr = list(map(int, stdin.readline().split()))
arr.sort()

freq = defaultdict(list)
for num in arr:
	count = 0
	while num > 0:
		freq[num].append(count)
		num //= 2
		count += 1
	freq[0].append(count)


# print(freq)

for i in freq:
	freq[i].sort()

ans = float('inf')
m = max(arr)
for i in range(0, m + 1):
	ans = min(ans, get_arr(i))
	
print(ans)