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:
def int_multiple():
    return  [int(c) for c in input().split()]

def int_single():
    return int(input())

def str_multiple():
    return [c for c in input().split()]

def str_single():
    return input()

# start

n, k = int_multiple()
l = int_multiple()

l = sorted(l)


costs = []
for i in range(200001):
    costs.append([])

for i in range(n):
    tmp = l[i]
    cnt = 0
    while (tmp != 0):
        costs[tmp].append(cnt)
        tmp = int(tmp/2)
        cnt += 1

for val in costs[1]:
    costs[0].append(val+1)


min_cost = 9999999999999
for c in costs:
    if len(c) >= k:
        cost = sum(c[:k])
        if (cost < min_cost):
            min_cost = cost

#for cc in costs:
#    print(cc)

print(min_cost)