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
def mp():
    return map(int, input().split())

def f(a, b):
    res = 0
    while b > a:
        b //= 2
        res += 1
    if b != a:
        res = -1
    return res

n, k = mp()
a = sorted(list(mp()))
m = max(a) + 1
b = [0] * m
c = [0] * m
for i in a:
    x = i
    cnt = 0
    while x > 0:
        if c[x] < k:
            b[x] += cnt
            c[x] += 1
        x //= 2
        cnt += 1
    if c[0] < k:
        b[0] += cnt
        c[0] += 1
ans = b[0]
for i in range(1, m):
    if c[i] == k:
        ans = min(ans, b[i])

print(ans)