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
n, k = map(int, input().split())
A = list(map(int, input().split()))
A.sort()

B = [0]*(2*10**5+1)
C = [0]*(2*10**5+1)
for i in range(n):
    a = A[i]
    cnt = 0
    while a:
        if C[a] < k:
            B[a] += cnt
            C[a] += 1
        a = a//2
        cnt += 1
    if C[a] < k:
        B[a] += cnt
        C[a] += 1

#print(B[0:10])
#print(C[0:10])
ans = 10**18
for i in range(len(C)):
    if C[i] >= k:
        ans = min(B[i], ans)
print(ans)
