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 = [int(i) for i in input().split()]
data = [int(i) for i in input().split()]

dics = []
for i in range(20):
    dics.append({})

for d in data:
    ln = len(bin(d)) - 2
    for i in range(ln, -1, -1):
        # print(bin(d))
        if d not in dics[i]:
            dics[i][d] = [ln - i]
        else:
            dics[i][d].append(ln - i)
        d >>= 1



mn = 1<<50
for i in range(20):
    dic = dics[i].values()
    for d in dic:
        if len(d) >= k:
            d.sort()
            val = sum(d[:k])
            if val < mn:
                mn = val


print(mn)

# ans2 = 0 
# vals = dics[ans].values()
# if len(vals) == 0:
#     pass
# else:
#     val = max(vals)
#     for d in dic