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 PYTHON):

Solution:
from collections import Counter,defaultdict
import heapq
from sys import stdin
raw_input = stdin.readline
n,k=map(int,raw_input().split())
l=map(int,raw_input().split())
d1=[0]*200005
d2=[0]*200005
d=[[] for i in xrange(200005)]
mx=0
for i in l:
    x=i
    d1[x]+=1
    heapq.heappush(d[x],0)
    if d1[x]>k:
        pp=-heapq.heappop(d[x])
        d2[x]-=pp
        d1[x]-=1
    c=0
    mx=max(mx,i)
    while x:
        x/=2
        c+=1
        heapq.heappush(d[x],-c)
        d1[x]+=1
        d2[x]+=c
        #print x,d1[x],k
        if d1[x]>k:
            pp=-heapq.heappop(d[x])
            d2[x]-=pp
            d1[x]-=1
        
ans=d2[0]
for i in xrange(mx+1):
    if d1[i]>=k:
        ans=min(ans,d2[i])
print ans