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:
def main():
    m, n = [int(v) for v in input().split()]
    vals = [int(v) for v in input().split()]
    data = {}
    data[0] = list()
    for v in vals:
        d = v
        count = 0
        while d != 0:
            if d not in data:
                data[d] = list()
            data[d].append(count)
            count += 1
            d = d // 2
        data[d].append(count)
    min_v = 1000000000
    for k, v in data.items():
        if len(v) >= n:
            v.sort()
            d = sum(v[0:n])
            if d < min_v:
                min_v = d

    print(min_v)


if __name__ == "__main__":
    main()