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()]
nums = sorted([int(x) for x in input().split()])
count = [0 for _ in range(2 * 100000 + 10)]
ops = [0 for _ in range(2 * 100000 + 10)]
for num in nums:
    op = 0
    while (num > 0):
        ops[num] += op if count[num] < k else 0
        count[num] += 1
        num = num // 2
        op += 1
    ops[num] += op if count[num] < k else 0
    count[num] += 1

minn = 0
minop = ops[0]
for i in range(2 * 100000 + 1):
    if (count[i] >= k and ops[i] < minop):
        minn = i
        minop = ops[i]

print(minop)