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
n, k = [int(x) for x in input().split()]
a = [int(x) for x in input().split()]
b = []
for i in range(n):
    b.append([a[i],0])
    a[i]//=2
    j=1
    while(a[i]>0):
        b.append([a[i],j])
        a[i]//=2
        j+=1
    b.append([a[i],j])
b = sorted(b, key = lambda x:(x[0],x[1]))
minop = 0
su = 0
for i in range(k):
    su += b[i][1]
minop = float('inf')
if(b[0][0]==b[k-1][0]):
    minop = su
for i in range(1,len(b)-k+1):
    if(b[i][0]!=b[i+k-1][0]):
        su += b[i+k-1][1]
        su -= b[i-1][1]
        continue
    su += b[i+k-1][1]
    su -= b[i-1][1]
    if(su<minop):
        minop = su
print(minop)