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:
"""
Code of Ayush Tiwari
Codechef: ayush572000
Codeforces: servermonk

"""
import sys
input = sys.stdin.buffer.readline

def solution():
    n,k=map(int,input().split())
    l=list(map(int,input().split()))
    l.sort()
    val=[0]*1000001
    op=[0]*1000001
    m=2e9
    for i in range(n):
        val[l[i]]+=1
    for i in range(n):
        cnt=0
        x=l[i]
        while x>0:
            if val[x]>=k:
                m=min(m,op[x])
            cnt+=1
            x//=2
            op[x]+=cnt
            val[x]+=1
            
    print(m)
solution()