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 PYTHON):

Solution:
from collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat
import math

# sys.stdin = open('input')

def inp():
    return map(int, raw_input().split())

def inst():
    return raw_input().strip()


def main():
        n, k = inp()
        da = inp()
        da.sort()
        ct = Counter(da)
        ans = int(1e20)
        ps = [0]
        for i in da:
            ps.append(ps[-1]+i)
        last = 0
        # print da
        # print ps
        for i in range(n):
            if da[i]==last:
                continue
            last = da[i]
            num = ct[last]
            if num>=k:
                print 0
                return
            cost_before = i*(da[i]-1) - ps[i]
            cost_after = ps[n] - ps[i+num] - (n-(i+num))*(da[i]+1)
            need = k-num
            ans = min(ans, cost_before+cost_after+need)
            if i>=need:
                ans = min(ans, cost_before+need)
            if n-(i+num)>=need:
                ans = min(ans, cost_after+need)
        print ans







main()