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 PYTHON):

Solution:
from collections import defaultdict
# from heapq import heappush, heappop
# from bisect import insort

n, k = [int(x) for x in raw_input().split()]
inputs = [int(x) for x in raw_input().split()]

def min_step(inputs, k):
    n = len(inputs)
    # reduced = [[] for _ in range(n)]

    reduce_count = defaultdict(int)
    reduce_step = defaultdict(lambda: [])
    for i in range(len(inputs)):
        x = inputs[i]
        count = 0
        while x > 0:
            reduce_count[x] += 1
            reduce_step[x].append(count)
            # reduced[i].append(x)
            x /= 2
            count += 1

    # print 'reduce_count', dict(reduce_count)
    # print 'reduce_step', dict(reduce_step)

    output = float("inf")
    didsort = defaultdict(lambda: False)
    for x in reduce_count:
        if reduce_count[x] >= k: # at least k numbers can reduce to x:
            current_count = 0
            
            if not didsort[x]:
                reduce_step[x].sort()
                didsort[x] = True
            # steps = reduce_step[x].values()
            # steps.sort()

            output = min(output, sum(reduce_step[x][:k]))

    return output

print(min_step(inputs, k))