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
"""
useful stuff 
x= list(map(int,input().split()))
x =[int (e) for e in input().split()]

swap x,y=y,x
substring a[x:y] mean x to y-1

print list
print(*list)
print(' '.join(map(str,listname)))

start from 1 to n
for i in range(1,n+1)

Don't froget dumb stuff like:
if: 
for:
while: 
and tab
def doesn't require data type
"""
z= []
a = []
for i in range (200010):
    a.append([])
n,k = [int (e) for e in input().split()]
b = [int (e) for e in input().split()]
for x in b:
    c=x
    ct=0
    a[x].append(0)
    while(c!=0):
        c>>=1
        ct+=1
        a[c].append(ct)
ans=10000000000
for i in range(200000):
    if(len(a[i])<k):
        continue
    a[i].sort()
    s=0
    for j in range(k):
        s+=a[i][j]
    ans=min(ans,s)
print(ans)
