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
# cook your dish here
import math
n,k=map(int,input().split(" "))
a=list(map(int,input().split()))
a=sorted(a)
m=max(a)//2
c=[a[0]]
d=[1]
for i in range(1,n):
    if a[i]==c[-1]:
        d[-1]+=1
    else:
        c.append(a[i])
        d.append(1)
#print(c,d)        
if max(d)>=k:
    print(0)
else:
    c=[[-1 for i in range(n)] for j in range(m+1)]
    s=[]
    i=n-1
    while i>=0:
        q=a[i]
        if q<=m:
            c[q][i]=0
        k1=1
        while q!=0:
            
            q=q//2
            c[q][i]=k1
            k1=k1+1
        i-=1
    for i in c:
        j=0
        k1=k
        count=0
        while j<n and k1>0:
            if i[j]!=-1:
                count+=i[j]
                k1-=1
            j+=1
        
        if k1==0:
            s.append(count)
    
    print(min(s))
            
            
        

