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=tuple(map(int,input().split()))
a=list(map(int,input().split()))
a1=sorted(a)
a=sorted(a)
l=[[0 for i in range(n)]for j in range(20)]
d={}
for j in range(n):
    x=1
    for i in range(20):
        if(i==0):
            l[i][j]=a[j]
        else:
            l[i][j]=a[j]//2 
            a[j]=a[j]//2

value=10**9

for i in range(20):
    for j in range(n):
        if(l[i][j]!=0):
            if(l[i][j] not in d):
                d[l[i][j]]=[1,i]
            else:
                if(d[l[i][j]][0]<k):
                    d[l[i][j]][0]+=1
                    d[l[i][j]][1]+=i 
for i in d:
    if(d[i][0]>=k):
        value=min(value,d[i][1])
#print(d)
temp_ans=0
for j in range(k):
    for i in range(1,20):
        x=2
        if(a1[j]//x  !=0):
            a1[j]=a1[j]//x
        else:
            temp_ans+=i-1
value=min(temp_ans,value)
print(value)