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:
import sys

n, k = map(int, sys.stdin.readline().split())

a = list(map(int, sys.stdin.readline().split()))

xs = [0]
for i in range(n):
    x = a[i]

    while x != 0:
        xs.append(x)
        x = x//2
xs = list(set(xs))

# op_cnts = [0]*(2*(10**5))
op_cnts = dict()
for x in xs:
    op_cnts[x] = []
# print(op_cnts)
for i in range(n):
    cnt = 0
    start = a[i]
    while True:

        if start in op_cnts:
            op_cnts[start].append(cnt)
        cnt += 1
        if start == 0:
            break
        start = start // 2
minVal = sys.maxsize

for x in xs:
    if len(op_cnts[x]) >= k:
        sumVal = 0
        op_cnts[x] = sorted(op_cnts[x])
        for j in range(k):
            sumVal += op_cnts[x][j]
        minVal = min(minVal, sumVal)

print(minVal)