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())
X = sorted(list(map(int, input().split())))
import sys;input=sys.stdin.readline
d = dict()
for x in X:
    if x not in d:
        d[x] = 0
    d[x] += 1
    if d[x] >= K:
        print(0)
        sys.exit()
L = [0]*N
b = X[0]
for i in range(1, N):
    L[i] = (X[i]-b)*i + L[i-1]
    b = X[i]

R = [0]*N
b = X[-1]
for i in range(N-2, -1, -1):
    R[i] = (b-X[i])*(N-1-i) + R[i+1]
    b = X[i]

r = 10**18
for i in range(N):
    cr = 10**18
    if i+1 >= K:
        cr = min(cr, L[i]-(i+1-K))
    if N-i >= K:
        cr = min(cr, R[i]-(N-i-K))
    if i+1 < K and N-i < K:
        cr = min(cr, L[i]+R[i]-(N-K))
#    print(i, cr)
    cr = max(cr, 0)
    r = min(cr, r)

#print(L, R)
print(r)