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
_, k = map(int, input().split())
arr = map(int, input().split())

number = dict()
A = list()

for n in arr:
    cnt = 0
    B = dict()
    while True:
        B[n] = cnt
        if n not in number:
            number[n] = 0
        number[n] += 1
        cnt += 1
        n //= 2
        if n == 0:
            break

    A.append(B)

best = 1_000_000_000

for n in range(0, 200_000):
    if n not in number or  number[n] < k:
        continue

    temp = []
    for m in A:
        if n not in m:
            continue

        temp.append(m[n])

    if len(temp) < k:
        break

    temp = sorted(temp)
    best = min(best, sum(temp[:k]))

print(best)

