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:
n,k=map(int,input().split())
a=list(map(int,input().split()))
dp=[[] for i in range(200005)]
mp={}
cnt=0
for i in a:
    if i not in mp:
        mp[i]=0
    else:
        mp[i] += 1
        cnt=max(cnt,mp[i])
if cnt >= k:
    print(0)
else:
    for i in a:
        cnt=0
        x=i
        while x:
            dp[x].append(cnt)
            x=x//2
            cnt += 1
        dp[x].append(cnt)
    ans=10000000000000000000000000000000000000000000000
    for i in range(200005):
        if len(dp[i])<k:
            continue
        else:
            cnt=0
            dp[i]=sorted(dp[i])
            for j in range(k):
                cnt += dp[i][j]
            ans=min(ans,cnt)
    print(ans)