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:
from collections import defaultdict,Counter
n,k = map(int,input().split())
List = [int(x) for x in input().split()]
List.sort()
count = Counter(List)
for i in count.keys():
    if(count[i]>=k):
        print(0)
        exit()
reached = [0] * 200001
cost = [0] * 200001
for i in range(n):
    val = List[i]
    curr_cost = 0
    while(val):
        val>>=1
        curr_cost += 1
        if(reached[val]<k-count[val]):
            cost[val] += curr_cost
            reached[val] += 1
ans = 10000000000
for i in range(200001):
    if(reached[i] == k-count[i]):
        ans = min(ans,cost[i])
print(ans)