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:
n, k = [int(p) for p in input().split()]
arr = [int(p) for p in input().split()]
arr.sort()
span = [-k]*(max(arr)+1)
cost = [0]*(max(arr)+1)
for i in range(n):
	j = 0
	while True:
		if span[arr[i]] != 0:
			span[arr[i]] += 1
			cost[arr[i]] += j
		arr[i] //= 2
		j += 1
		if arr[i] == 0:
			if span[arr[i]] != 0:
				span[arr[i]] += 1
				cost[arr[i]] += j
			break
ans = float('inf')
for i in range(len(span)):
	if span[i] == 0:
		ans = min(ans, cost[i])
print(ans)