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=map(int,input().split())
a=input().split()
for i in range(n):
    a[i]=int(a[i])
a.sort()
a1=[]
for j in range(n):
    a1.append(a[j])
b=[0 for i in range(2*(10**5)+1)]
b1=[0 for i in range(2*(10**5)+1)]
for j in range(n):
    q=0
    while a[j]:
        if(b[a[j]]<k):
            b[a[j]]+=1
            b1[a[j]]+=q
        a[j]=a[j]//2  
        q+=1
c=[]        
for j in range(2*(10**5)+1):
    if(b[j]==k):
        c.append(b1[j])
count=0    
#print(c)
#print(a1)
for i in range(k):
    while a1[i]!=0:
        count+=1
        a1[i]=a1[i]//2
#print(count,min(c))        
if(len(c)>0):print(min(count,min(c)))  
else:print(count)