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 PYTHON3):

Solution:
n,k=map(int,input().split())
x=list(map(int,input().split()))
x.sort()
dp=[0]*n
ma,count=0,1
for i in range(n):
    if i>0 and x[i]==x[i-1]:
        count+=1
        ma=max(ma,count)
    else:
        count=1
    if i==0:dp[i]=x[i]
    else: dp[i]=dp[i-1]+x[i]
ans=123456789999990
for i in range(1,n):
    if i+1>=k:
        ans=min(ans,x[i]*i-dp[i-1]-(i+1-k))
    if n-i>=k:
        ans = min(ans,dp[n - 1]-dp[i]-x[i]*(n-i-1)-(n-i-k))
    ans=min(ans,x[i]*i-dp[i-1]+dp[n-1]-dp[i]-x[i]*(n-1-i)-(n-k))
if ma>=k:ans=0
if n==1:ans=0
print(max(ans,0))