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 
import sys
range = xrange
input = raw_input

inp = [int(x) for x in sys.stdin.read().split()]; ii = 0

n = inp[ii]; ii += 1
m = inp[ii]; ii += 1

X = inp[ii: ii + n]; ii += n

marked = set()
for x in X:
    marked.add(x)
marked.add(-2*10**9 - 1)
marked.add( 2*10**9 + 1)

pos = []
for x in X:
    if x + 1 not in marked:
        pos.append(x + 1)
        marked.add(x + 1)
    if x - 1 not in marked:
        pos.append(x - 1)
        marked.add(x - 1)
pick = []
tmp = []
cost = 1
total = 0
while m:
    pos.reverse()
    while m and pos:
        m -= 1
        total += cost
        x = pos.pop()
        pick.append(x)

        if x + 1 not in marked:
            tmp.append(x + 1)
            marked.add(x + 1)
        if x - 1 not in marked:
            tmp.append(x - 1)
            marked.add(x - 1)
    tmp,pos = pos,tmp
    cost += 1

print total
print ' '.join(str(x) for x in pick)
