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:
import sys
input = sys.stdin.readline

n, k = [int(i) for i in input().split()]
a = [int(i) for i in input().split()]
a.sort()

for i in range(n-k+1):
    if a[i] == a[i+k-1]:
        print(0)
        sys.exit(0)

l = r = m = 0
# suppose val is the number where k numbers are the same
# case 1, turn the leftmost k numbers to val (or val-1)
for i in range(k-1):
    l += a[k-1] - a[i]
for i in range(k-1, n-1):  # if a[k-1] have many same numbers
    if a[i] == a[i+1]:
        l -= 1
    else:
        break

# case 2, turn the rightmost k numbers to val (or val-1)
a_reverse = a[::-1]
for i in range(k-1):
    r += -(a_reverse[k-1] - a_reverse[i])
for i in range(k-1, n-1):
    if a_reverse[i] == a_reverse[i+1]:
        r -= 1
    else:
        break

# case 3, all the numbers are val or val-1 or val+1
for i in range(n//2):
    m += a[n-i-1] - a[i]
m -= n - k
print(min(l, r, m))