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
from collections import deque
n,k=map(int,input().split())
a=list(map(int,input().split()))
counted=[0 for i in range(200001)]
stored=[0 for i in range(200001)]
a.sort(reverse=True)
queue=deque()
flag=0
for i in range(n):
    stored[a[i]]+=1
    if(stored[a[i]]==k):
        print(0)
        flag=1
        break
if(flag==0):
    for i in range(n):
        queue.append([a[i],0])
    while(queue):
        num,count=queue.popleft()
        if(stored[num//2]<k):
            counted[num//2]+=count+1
            stored[num//2]+=1
        if(num//2>0):
            queue.append([num//2,count+1])
    mini=99999999
    for i in range(200001):
        if(stored[i]==k):
            mini=min(mini,counted[i])
    print(mini)