There are n Christmas trees on an infinite number line. The i-th tree grows at the position x_i. All x_i are guaranteed to be distinct.

Each integer point can be either occupied by the Christmas tree, by the human or not occupied at all. Non-integer points cannot be occupied by anything.

There are m people who want to celebrate Christmas. Let y_1, y_2, ..., y_m be the positions of people (note that all values x_1, x_2, ..., x_n, y_1, y_2, ..., y_m should be distinct and all y_j should be integer). You want to find such an arrangement of people that the value ∑_{j=1}^{m}min_{i=1}^{n}|x_i - y_j| is the minimum possible (in other words, the sum of distances to the nearest Christmas tree for all people should be minimized).

In other words, let d_j be the distance from the j-th human to the nearest Christmas tree (d_j = min_{i=1}^{n} |y_j - x_i|). Then you need to choose such positions y_1, y_2, ..., y_m that ∑_{j=1}^{m} d_j is the minimum possible.

Input

The first line of the input contains two integers n and m (1 ≤ n, m ≤ 2 ⋅ 10^5) — the number of Christmas trees and the number of people.

The second line of the input contains n integers x_1, x_2, ..., x_n (-10^9 ≤ x_i ≤ 10^9), where x_i is the position of the i-th Christmas tree. It is guaranteed that all x_i are distinct.

Output

In the first line print one integer res — the minimum possible value of ∑_{j=1}^{m}min_{i=1}^{n}|x_i - y_j| (in other words, the sum of distances to the nearest Christmas tree for all people).

In the second line print m integers y_1, y_2, ..., y_m (-2 ⋅ 10^9 ≤ y_j ≤ 2 ⋅ 10^9), where y_j is the position of the j-th human. All y_j should be distinct and all values x_1, x_2, ..., x_n, y_1, y_2, ..., y_m should be distinct.

If there are multiple answers, print any of them.

Examples

Input


2 6
1 5


Output


8
-1 2 6 4 0 3 


Input


3 5
0 3 1


Output


7
5 -2 4 -1 2 
def arr_inp():
    return [int(x) for x in stdin.readline().split()]


def bfs():
    global m
    ans, queue, dis = [], deque([]), 0

    for i in a:
        visit[i] = 1
        queue.append(i)

    while queue:
        # dequeue parent vertix
        s = queue.popleft()

        # enqueue child vertices
        for i in [s - 1, s + 1]:
            if m:
                if visit[i] == 0:
                    queue.append(i)
                    ans.append(i)
                    m -= 1
                    visit[i] = 1
            else:
                queue = deque([])
                break
    return ans


from sys import stdin
from collections import defaultdict, deque
from bisect import *

n, m = arr_inp()
a, visit, ans = sorted(arr_inp()), defaultdict(int), 0
arr = bfs()

for i in arr:
    ix = bisect_right(a, i)
    if ix == 0:
        ans += abs(a[0] - i)
    elif ix == n:
        ans += abs(a[-1] - i)
    else:
        ans += min(abs(a[ix - 1] - i), abs(a[ix] - i))

print(ans)
print(*arr)
