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 = map(int, input().split())

a_list = sorted(map(int, input().split()))

nums = [0 for i in range(1 + 2* (10**5))]
counts = [0 for i in range(1 + 2* (10**5))]

for a in a_list:
    count = 0
    nums[a] += 1
    while a != 0:
        a = a // 2
        count += 1
        if nums[a] < k:
            nums[a] += 1
            counts[a] += count
        #elif nums[a] == k:
        #    break

#print(nums[:10])
#print(counts[:10])

counts2 = [counts[i] for i, num in enumerate(nums) if num >= k]
out = min(counts2)
print(out)