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())
res = k*20
a = list(map(int, input().split()))

ansNum = dict()
ansCost = dict()
ansCost.fromkeys(a, 0)
for i in a:
    ansNum[i] = ansNum.get(i, 0) + 1
    if ansNum[i] >= k:
        res = 0


l = n
c = 0
while True:
    ir = 0
    iw = 0
    c+=1
    if l <= 0:
        break

    while iw < l:
        if a[ir] == 0:
            l-=1
        else:
            a[iw] = a[ir]>>1
            ansNum[a[iw]] = ansNum.get(a[iw], 0)+1
            ansCost[a[iw]] = ansCost.get(a[iw], 0)+c
            if ansNum[a[iw]] == k:
                res = min(res, ansCost[a[iw]])
            iw+=1
        ir+=1

print(res)