1213_D2. Equalizing by Division (hard version)
The only difference between easy and hard versions is the number of elements in the array.

You are given an array a consisting of n integers. In one move you can choose any a_i and divide it by 2 rounding down (in other words, in one move you can set a_i := ⌊(a_i)/(2)⌋).

You can perform such an operation any (possibly, zero) number of times with any a_i.

Your task is to calculate the minimum possible number of operations required to obtain at least k equal numbers in the array.

Don't forget that it is possible to have a_i = 0 after some operations, thus the answer always exists.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the number of elements in the array and the number of equal numbers required.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 2 ⋅ 10^5), where a_i is the i-th element of a.

Output

Print one integer — the minimum possible number of operations required to obtain at least k equal numbers in the array.

Examples

Input


5 3
1 2 2 4 5


Output


1


Input


5 3
1 2 3 4 5


Output


2


Input


5 3
1 2 3 3 3


Output


0
```python
from sys import stdin

n, k = [int(i) for i in stdin.readline().strip().split()]
a = [int(i) for i in stdin.readline().strip().split()]


p = [19 * [0] for _ in range(max(a) + 1)]


nums = set()
for m in a:
    for i in range(19):
        p[m >> i][i] += 1
        nums.add(m >> i)

        if (m >> i) == 0:
            break

res = 10000000000000000000

for m in nums:
    d = p[m]
    sum = 0
    steps = 0
    for i in range(19):
        if sum + d[i] >= k:
            add = (k - sum) * i
            steps += add
            res = min(steps, res)
            break
        else:
            sum += d[i]
            steps += (i * d[i])

print(res)

```