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 = map(int, input().split())
A = list(map(int, input().split()))
G = []
for _ in range(2 * 10 ** 5 + 100):
    G.append([])
c = 0
for i in A:
    if len(G[i]) < k:
        G[i].append(0)
while True:
    c += 1
    f = True
    for i in range(n):
        if A[i]:
            A[i] //= 2
            if len(G[A[i]]) < k:
                G[A[i]].append(c)
        if A[i]:
            f = False
    if f:
        break
ans = 10 ** 9
for t in G:
    if len(t) == k:
        ans = min(ans, sum(t))
print(ans)