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;
void _main(int tc) {
  int n, m;
  cin >> n >> m;
  deque<pair<int, int> > q;
  for (int i = 0; ((0 < n ? 1 : -1) < 0) ? i > n : i < n;
       i += (0 < n ? 1 : -1)) {
    int x;
    cin >> x;
    q.push_back({x, 0});
  }
  set<int> vis;
  for (int i = 0; ((0 < int(q.size()) ? 1 : -1) < 0) ? i > int(q.size())
                                                     : i < int(q.size());
       i += (0 < int(q.size()) ? 1 : -1)) {
    vis.insert(q[i].first);
  }
  long long int sumAns = 0;
  vector<int> ans;
  while (int(q.size())) {
    auto u = q.front();
    q.pop_front();
    if (!vis.count(u.first + 1)) {
      vis.insert(u.first + 1);
      q.push_back({u.first + 1, u.second + 1});
      ans.push_back(u.first + 1);
      sumAns += u.second + 1;
    }
    if (int(ans.size()) == m) break;
    if (!vis.count(u.first - 1)) {
      vis.insert(u.first - 1);
      q.push_back({u.first - 1, u.second + 1});
      ans.push_back(u.first - 1);
      sumAns += u.second + 1;
    }
    if (int(ans.size()) == m) break;
  }
  cout << sumAns << '\n';
  for (int i = 0; ((0 < m ? 1 : -1) < 0) ? i > m : i < m;
       i += (0 < m ? 1 : -1)) {
    cout << ans[i] << " \n"[i + 1 == m];
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  _main(0), exit(0);
  int tc;
  cin >> tc;
  for (int i = 0; ((0 < tc ? 1 : -1) < 0) ? i > tc : i < tc;
       i += (0 < tc ? 1 : -1))
    _main(i + 1);
}
