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
# import io, os
# input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
import sys
# sys.stdin=open('input.txt','r')
# sys.stdout=open('output.txt','w')
input=sys.stdin.readline
# sys.setrecursionlimit(300010)
MOD = 1000000007
MOD2 = 998244353
ii = lambda: int(input().strip('\n'))
si = lambda: input().strip('\n')
dgl = lambda: list(map(int,input().strip('\n')))
f = lambda: map(int, input().strip('\n').split())
il = lambda: list(map(int, input().strip('\n').split()))
ls = lambda: list(input().strip('\n'))
lsi = lambda: [int(i) for i in ls()]
let = 'abcdefghijklmnopqrstuvwxyz'
for _ in range(1):
    n,k=f()
    l=il()
    mx=max(l)
    cost=[[] for i in range(mx+1)]
    for i in l:
        x=0
        while i>0:
            cost[i].append(x)
            x+=1
            i//=2
        cost[0].append(x)
    mn=10**15
    for i in range(mx+1):
        if len(cost[i])>=k:
            cost[i].sort()
            mn=min(mn,sum(cost[i][:k]))
    print(mn)