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 = [int(i) for i in input().split()]
a.sort()
data = {}
for x in a:
    cur = x
    iterations = 0
    while (cur > 0):
        if data.get(cur) is not None:
            data[cur].append(iterations)
        else:
            data[cur] = [iterations]
        cur = cur // 2
        iterations += 1
    if data.get(0) is not None:
        data[0].append(iterations)
    else:
        data[0] = [iterations]
ans = 100000000000
for key, iters in data.items():
    if (len(iters) >= k):
        s = 0
        for i in range(k):
            s += iters[i]
        ans = min(ans, s)
print(ans)