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())
l=list(map(int,input().split()))
l.sort()
ld=[]
lj=[]
le=[]
for i in range(2*10**5+1):
    h=[0,0]
    le.append(h)
for i in range(n):
    g=[]
    j=0
    while(l[i]):
        g.append(l[i])
        l[i]=l[i]//2
        j+=1
    ld.append(g)
    lj.append(j)
j=0
mi=10**18
for i in range(n):
    for j in range(lj[i]):
        if le[ld[i][j]][0]<k:
            le[ld[i][j]][0]+=1
            le[ld[i][j]][1]+=j
        if le[ld[i][j]][0]==k:
            if le[ld[i][j]][1]<mi:
                mi=le[ld[i][j]][1]
print(mi)