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:
# 7=>3=>1, 2=>1
# 8/9
# 4=>2=>1

def cal_dis(num,val):
    count = 0
    while num > 0:
        if val[num] == 0: val[num] = []
        val[num].append(count)
        num = num//2
        count += 1
    return count,val

n,k = (int(x) for x in input().split())
seq = [int(x) for x in input().split()]
ans = 10e8
val = 200009 * [0]
for ele in seq:
    count,val = cal_dis(ele,val)

for lis in val:
    if lis != 0:
        if lis.__len__() >= k:
            lis.sort()
            sum = 0
            for ele in lis[:k]:
                sum += ele
            ans = min(ans,sum)
print(ans)