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 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()))
b = set()
for i in a:
    x = i
    while x > 0:
        b.add(x)
        x //= 2
b.add(0)

tot = 10 ** 20
for x in b:
    ans = cnt = 0
    for i in range(n):
        if a[i] == x:
            cnt += 1
        elif a[i] > x:
            res = f(x, a[i])
            if res != -1:
                ans += res
                cnt += 1
        if cnt >= k:
            break
    if cnt >= k and ans < 10 ** 20:
        tot = min(tot, ans)
        #print(x, ans)

print(tot)