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
input = sys.stdin.readline

n, k = list(map(int, input().split()))
A = list(map(int, input().split()))
A.sort()

ops = [0] * 200007

for i in range(n):
    alph = A[i]
    while alph > 0:
        ops[alph] += 1
        alph //= 2

    ops[0] += 1

ans = float('inf')
for i in range(200007):
    if ops[i] >= k:
        count = 0
        numOps = 0
        for j in range(n):
            if count == k:
                break
            alph = A[j]
            tmpOps = 0
            while alph > i:
                alph //= 2
                tmpOps += 1
            if alph == i:
                count += 1
                numOps += tmpOps

        ans = min(ans, numOps)

print(ans)