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;
int main() {
  pair<long long int, long long int> v;
  long long int n, m, i, j, ans = 0;
  queue<long long int> q;
  cin >> n >> m;
  vector<long long int> res, a(n);
  vector<pair<long long int, long long int> > d(n + 2);
  for (i = 1; i <= n; i++) {
    cin >> a[i - 1];
  }
  sort(a.begin(), a.end());
  v.second = -2e9 - 1;
  d[0] = v;
  v.first = 2e9 + 1;
  d[n + 1] = v;
  for (i = 1; i <= n; i++) {
    d[i].first = a[i - 1];
    d[i].second = a[i - 1];
    q.push(i);
  }
  i = 0;
  while (i < m) {
    j = q.front();
    q.pop();
    bool f = 0;
    if (d[j].second + 1 < d[j + 1].first && i < m) {
      ans += d[j].second + 1 - a[j - 1];
      res.push_back(d[j].second + 1);
      d[j].second++;
      i++;
      f = 1;
    }
    if (d[j].first - 1 > d[j - 1].second && i < m) {
      ans += a[j - 1] - d[j].first + 1;
      res.push_back(d[j].first - 1);
      d[j].first--;
      i++;
      f = 1;
    }
    if (f) q.push(j);
  }
  cout << ans << '\n';
  for (i = 0; i < res.size(); i++) cout << res[i] << " ";
  return 0;
}
