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:
n, k = tuple(map(int, input().split()))
arr = list(map(int, input().split()))
arr.sort()

karms = []
for i in arr:
    t = []
    while i > 0:
        t.append(i)
        i //= 2
    t.append(0)
    karms.append(t)

#print(*karms)

ans = 1e15
for i in range(0, arr[-1] + 1):
    count = 0
    for karm in karms:
        if i in karm:
            count += 1
        if count == k:
            break

    if count < k:
        continue
    adds = []
    for karm in karms:
        if i not in karm:
            continue
        adds.append(karm.index(i))
        if len(adds) == k:
            break
    ans = min(ans, sum(adds))

print(ans)