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 convert_to_bin(x):
    res = ""
    while x>0:
        if x%2 == 0:
            res += "0"
        else:
            res += "1"
        x = x//2
    return res[::-1]

in_file =sys.stdin #  open("D.txt", "r") # 

n, k = map(int, in_file.readline().strip().split())
A = list(map(int, in_file.readline().strip().split()))
A = [convert_to_bin(a) for a in A]
lookup = {}
kids = []
parent = []
count = []
best = []
for a in A:
    t = ""
    for i in range(len(a)):
        prev = t
        t += a[i]
        if t not in lookup:
            lookup[t] = len(kids)
            if i != 0:
                kids[lookup[prev]].append(len(kids))
                parent.append(lookup[prev])
            else:
                parent.append(-1)
            kids.append([])
            
            count.append(0)
            best.append([-1 for i in range(k+1)])
            best[-1][0] = 0
        if i == len(a)-1:
            count[lookup[t]] += 1

def dp(i, k):
    if k < 0:
        return 0
    if best[i][k] != -1:
        return best[i][k]
    if count[i]>=k:
        best[i][k] = 0
        return 0
    best[i][k] = (len(A)+1)*(len(A)+1)*(len(A)+1)
    if len(kids[i]) == 1:
        best[i][k] = k-count[i] + dp(kids[i][0], k-count[i])
    elif len(kids[i]) == 2:
        for j in range(k-count[i]+1):
            best[i][k] = min(best[i][k], k-count[i]+ dp(kids[i][0], j)+dp(kids[i][1], k-count[i]-j))
    return best[i][k]

m = (len(A)+1)*(len(A)+1)*(len(A)+1)
for i in range(len(kids)):
    if m > dp(i, k):
        m = dp(i, k)


sys.stdout.write(str(m))

            

sys.stdout.flush()