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
ans = {}
n,k = (int(s) for s in input().split())
l = [int(s) for s in input().split()]
l.sort()

for i in range(n):
    x = l[i]
    count = 0
    while (x>0):
        if x in ans:
            if ans[x][0] <k:
                ans[x][0]+=1
                ans[x][1]+=count
        else:
            ans[x] = [1, count]
        count+=1
        x = x//2
    if 0 in ans:
        if ans[0][0] < k:
            ans[0][0] += 1
            ans[0][1] += count
    else:
        ans[x] = [1, count]
# print(ans)
ansm = 1000000000007
for i in ans:
    if ans[i][0]>=k:
        ansm = min(ansm, ans[i][1])
print(ansm)
