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()))

def check(x,y):
    if x>y:
        return -1
    cnt = 0
    while x<y:
        cnt+=1
        y = y//2
    if x==y:
        return cnt
    else:
        return -1

def find(x):
    ans = []
    for i in range(n):
        c = check(x,a[i])
        if c!=-1:
            ans += [c]
    if len(ans)>=k:
        # print(ans)
        return sum(sorted(ans)[:k])
    else:
        return -1

l = 0
r = 2*10**5+1
# r = 10
ret = 10**18
# while l<r:
#     md = (l+r+1)//2
#     v = find(md)
#     print(md,v)
#     if v!=-1:
#         ret = min(ret,v)
#         l = md
#     else:
#         r = md-1

for i in range(r):
        v = find(i)
        if v!=-1:
            ret = min(ret,v)

# print(l,find(4))
print(ret)