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 
#include <bits/stdc++.h>
using namespace std;
const int N = 2E5 + 10;
const int inf = 1 << 30;
long long read() {
  long long a = 0;
  char b = 1, c;
  do
    if ((c = getchar()) == 45) b = -1;
  while (c < 48 || c > 57);
  do a = (a << 3) + (a << 1) + (c & 15);
  while ((c = getchar()) > 47 && c < 58);
  return a * b;
}
void write(long long x, char c) {
  if (x < 0) putchar('-'), x = -x;
  char a[20], s = 0;
  do a[++s] = x % 10 | 48;
  while (x /= 10);
  do putchar(a[s]);
  while (--s);
  putchar(c);
}
int a[N], b[N];
set<int> vis;
queue<pair<int, int> > q;
int main() {
  int n = read(), m = read();
  for (int i = 1; i <= n; ++i) {
    a[i] = read();
    vis.insert(a[i]);
    q.push({a[i], -1});
    q.push({a[i], 1});
  }
  long long sum = 0;
  for (int s = 0; s < m;) {
    pair<int, int> x = q.front();
    q.pop();
    int k = x.second, w = x.first + k;
    if (vis.count(w)) continue;
    vis.insert(w);
    b[++s] = w;
    if (k > 0) {
      sum += k;
      q.push({x.first, k + 1});
    } else {
      sum -= k;
      q.push({x.first, k - 1});
    }
  }
  write(sum, '\n');
  for (int i = 1; i <= m; ++i) write(b[i], ' ');
}
