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:
l=input().split()
n=int(l[0])
k=int(l[1])
hashi=dict()
l=input().split()
li=[int(i) for i in l]
z=0
for i in li:
    z=i
    count=0
    while(z>0):
        if(z not in hashi):
            hashi[z]=[count]
        else:
            hashi[z].append(count)
        z=z//2
        count+=1
    if(0 not in hashi):
        hashi[0]=[count]
    else:
        hashi[0].append(count)
maxa=10**18
#print(hashi)
for i in hashi:
    if(len(hashi[i])<k):
        continue
    arr=[0 for i in range(30)]
    for j in hashi[i]:
        arr[j]+=1
    ans=0
    z=k
    for i in range(30):
        if(z<=arr[i]):
            ans=ans+(z*i)
            break
        ans=ans+(arr[i]*i)
        z-=arr[i]
    maxa=min(maxa,ans)
print(maxa)