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:
n,k=map(int,input().split())
a=list(map(int,input().split()))
v=[[] for _ in range(5*10**5+5)]
for i in range(n):
    cnt=0
    while a[i]!=0:
        v[a[i]].append(cnt)
        a[i]//=2
        cnt+=1
    v[0].append(cnt)
#for i in range(6):
    #print(v[i])
minc=10**11
for i in range(5*10**5+5):
    #print(len(v[i]))
    if(len(v[i])>=k):
        v[i].sort()
        #print(sum(v[i][:k]))
        minc=min(minc,sum(v[i][:k]))
#print(v)
print(minc)