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(x) for x in input().split()]
l = [int(x) for x in input().split()]
lk = [0] * (2 * 10**5 + 1)
ts = []
for i in range(2 * 10**5 + 1): ts.append([9999999])
for n in l:
	t = n
	c = 0
	while True:
		if(t == 0): break
		lk[t] += 1
		ts[t].append(c)
		t //= 2
		c += 1
mn = 9999999
for i in range(len(lk) - 1, -1, -1):
	if(lk[i] >= K):
		ts[i].sort()
		mn = min(mn, sum(ts[i][:K]))
print(mn)