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:
import sys
def i():
    return sys.stdin.readline()[:-1]

def count(l,i):
    return sum(x==i for x in i)
l,desiredSame = map(int,i().split())
nums = list(map(int,i().split()))

pos = set()
pos.add(0)
for item in nums:
    while item > 0:
        pos.add(item)
        item >>=1

currMin = 9000
for x in pos:
    distance = []
    for item in nums:
        shift = 0
        while item > x:
            shift += 1
            item >>= 1
        if item == x:
            distance.append(shift)
    if len(distance) >= desiredSame:
        currMin = min(currMin, sum(sorted(distance)[:desiredSame]))
print(currMin)