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;
long long int bg = 9982443530987;
long long int power(long long int x, long long int y) {
  if (x == 0) return 0;
  if (y == 0) return 1;
  long long int s = y / 2;
  long long int f = power(x, s);
  if (y % 2 == 0)
    return (f * f);
  else
    return f * f * x;
}
void boom() {
  long long int i, n, h, j, m;
  cin >> n >> m;
  long long int arr[n];
  set<pair<long long int, long long int> > set1;
  set<long long int> f;
  for (i = 0; i < n; i++) {
    cin >> arr[i];
    f.insert(arr[i]);
  }
  sort(arr, arr + n);
  for (i = 0; i < n; i++) {
    if (f.find(arr[i] - 1) == f.end()) set1.insert(make_pair(1, arr[i] - 1));
    if (f.find(arr[i] + 1) == f.end()) set1.insert(make_pair(1, arr[i] + 1));
  }
  set<long long int> vec1;
  long long int sum = 0;
  while (set1.size() != 0) {
    pair<long long int, long long int> r = *(set1.begin());
    set1.erase(make_pair(r.first, r.second));
    if (f.find(r.second) == f.end()) {
      sum += r.first;
      vec1.insert(r.second);
      long long int k = r.second;
      f.insert(r.second);
      if (f.find(k - 1) == f.end())
        set1.insert(make_pair(r.first + 1, k - 1));
      else if (f.find(k + 1) == f.end())
        set1.insert(make_pair(r.first + 1, k + 1));
      if (vec1.size() == m) break;
    }
  }
  cout << sum << '\n';
  for (set<long long int>::iterator d = vec1.begin(); d != vec1.end(); d++)
    cout << *d << " ";
}
int main() {
  long long int top;
  top = 1;
  while (top--) boom();
  return 0;
}
