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())
s=[int(x) for x in input().split()]
kk=[]
uni=dict()
for i in range(0,len(s)):
    N=2*s[i]
    c=-1
    while(N>0):
        c+=1
        N=N//2
        if(uni.get(N)==None):
            uni[N]=[c]
            kk.append(N)
        else:
            uni[N].append(c)
        
ans=20*n
for i in kk:
    l=uni[i]
    if(len(l)>=k):
        l=sorted(l)
        sumu=0
        for j in range(k):
            sumu+=l[j]

        ans=min(ans,sumu)

print(ans)