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
def find(A,K):
    counting=[[] for i in range(max(A)+1)]
    okay=[0]*(max(A)+1)
    for i in range(len(A)):
        count=0
        temp=A[i]
        
        while temp>0:
            counting[temp]+=[count]
            okay[temp]+=1
            temp=temp//2
            count+=1
        counting[temp]+=[count]
        okay[temp]+=1

    MIN=float("inf")
    for i in range(len(okay)):
        if okay[i]>=K:
            counting[i]=sorted(counting[i])
            MIN=min(MIN,sum(counting[i][:K]))
    #print(counting)
    #print(okay)
    return MIN
    
n,K=list(map(int,input().strip().split(" ")))
A=list(map(int,input().strip().split(" ")))
print(find(A,K))
    