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
def main():
    def input():
        return sys.stdin.readline()[:-1]
    n, l = map(int,input().split())
    a = list(map(int,input().split()))
    ans = 10000
    for e in range(0,max(a)+1):
        d = []
        t = 0
        for k in range(n):
            s = 0
            b = 0
            if a[k] == e:
                d.append(0)
            elif a[k] > e:
                b = a[k]
                while b > e:
                    b //= 2
                    s += 1
                if b == e:
                    d.append(s)
                else:
                    d.append(10000)
            elif a[k] < e:
                d.append(10000)
        ans = min(ans,sum(sorted(d)[:l]))
    print(ans)
if __name__ == '__main__':
    main()