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
# Why do we fall ? So we can learn to pick ourselves up.
def solve():
    n,k = map(int,input().split())
    aa = sorted([int(i) for i in input().split()])
    cc = [[] for _ in range(0,max(aa)+1)]
    for i in aa:
        ii = 0
        while i > 0:
            cc[i].append(ii)
            i //= 2
            ii += 1
        cc[i].append(ii)
    maxx = 10**9+7
    for i in cc:
        if len(i) >= k:
            maxx = min(maxx,sum(i[:k]))
    print(maxx)
solve()