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 = sorted(list(map(int, input().split())))
sums, suf, q1 = [0]*n, [n]*n, n
for q in range(n-2, -1, -1):
    sums[q] = sums[q + 1]
    if a[q] != a[q1-1]:
        sums[q] += (n-q1)*(a[q1-1]-a[q])+(q1-q-1)*(a[q1-1]-a[q]-1)
        q1 = q + 1
    suf[q] = q1
ans, sum1, q1 = sums[0]+k-suf[0] if suf[0] < k else 0, 0, -1
for q in range(1, n):
    if a[q] != a[q1 + 1]:
        sum1 += (q1+1)*(a[q]-a[q1+1])+(q-q1-1)*(a[q]-a[q1+1]-1)
        q1 = q - 1
    if suf[q]-q1-1 >= k:
        ans = 0
        break
    if k <= suf[q]:
        ans = min(ans, sum1+k-suf[q]+q1+1)
    if k < n-q1:
        ans = min(ans, sums[q]+k-suf[q]+q1+1)
    ans = min(ans, sum1+sums[q]+k-suf[q]+q1+1)
print(ans)