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
# t=int(input())
t=1
for j in range(t):
    n,k=list(map(int,input().strip().split()))
    a=list(map(int,input().strip().split()))
    arr=[]
    for i in range(2*10**5+2):
        arr.append([])
    for i in a:
        x=0
        xx=i
        arr[xx].append(x)
        while(xx):
            xx=xx//2
            x+=1
            arr[xx].append(x)
    op=sum(arr[0])
    # print(arr)
    for i in arr:
        
        if(len(i)<k):
            continue
        x=i.copy()
        x.sort()
        op=min(op,sum(x[:k]))
    print(op)
        