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 debug import debug
import sys, bisect
from math import log2 as log
mod = int(1e9)+7
inf = int(1e10)
input = sys.stdin.readline

n, k = map(int, input().split())
lis = sorted(map(int, input().split()))
last = int(2e5)+1
answer = inf
for i in range(1, last):
	x, y = bisect.bisect_left(lis, i), bisect.bisect_right(lis, i)
	f = y-x
	ans = 0
	for j in range(y, n): 
		tt = lis[j]
		c = 0
		while tt != 1:
			if tt == i: break
			tt = tt//2
			c+=1
		if tt == i: f += 1; ans += c
		if f>=k: break
	if f>=k:
		answer = min(answer, ans)
print(answer)
