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 sys import stdin 
n,k=map(int,stdin.readline().split())
l=[int(i) for i in stdin.readline().split()]
from collections import Counter
c=Counter(l)
if max(c.values())>=k:
    print(0)
    exit() 
from collections import defaultdict 
cc={}
for i in sorted(c.keys()):
    cc[i]=c[i]
d2=defaultdict(int)
f=defaultdict(int)
req=defaultdict(int)
for i in sorted(c.keys()):
    curr=i 
    ind=0 
    while curr:
        now=d2[curr]
        if now+c[i]<=k:
            d2[curr]+=cc[i]
            req[curr]+=ind*cc[i]
        else:
            f[curr]=1 
            d2[curr]=k 
            req[curr]+=(k-now)*ind 
        curr//=2
        ind+=1 
    now=d2[curr]
    if now+c[i]<=k:
        d2[curr]+=cc[i]
        req[curr]+=ind*cc[i]
    else:
        f[curr]=1 
        d2[curr]=k 
        req[curr]+=(k-now)*ind 
mini=10**9 
#print(d2)
#print(req)
for i in d2:
    if d2[i]<k:
        continue 
    mini=min(mini,req[i])
print(mini)