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:



from collections import Counter
n,k = map(int,input().split())
a = sorted(map(int,input().split()))
arr = Counter(a)
lsum = 0
lcnt = 0
rsum= sum(a)
rcnt = len(a)
ans = 10**18
for val,cnt in sorted(arr.items()):
    req = max(0,k-cnt)
    if(req==0):
        ans = 0
        break
    rcnt = rcnt-cnt
    rsum = rsum-val*cnt
    lr = min(req,lcnt)
    rr = max(0,req-lr)
    ta = ((val-1)*lcnt) - lsum + lr
    if lr<req:
        ta += (rsum-((val+1)*rcnt)) + rr
    ans = min(ans,ta)
    rr = min(rcnt,req)
    lr = max(0,req-rr)
    ta = rsum - ((val+1)*rcnt)+ rr
    if rr<req:
        ta += ((val-1)*lcnt) - lsum + lr
    ans = min(ans,ta)
    lcnt += cnt
    lsum += cnt*val
print(ans)