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>
#pragma GCC optimize("O3")
using namespace std;
vector<long long> bfs(queue<long long> q, set<long long> vis, int m) {
  vector<long long> res;
  while (true) {
    int first = q.front();
    q.pop();
    if (vis.find(first - 1) == vis.end()) {
      vis.insert(first - 1);
      res.push_back(first - 1);
      q.push(first - 1);
      m--;
    }
    if (m == 0) break;
    if (vis.find(first + 1) == vis.end()) {
      vis.insert(first + 1);
      res.push_back(first + 1);
      q.push(first + 1);
      m--;
    }
    if (m == 0) break;
  }
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  vector<int> arr(n);
  queue<long long> q;
  set<long long> vis;
  for (int i = 0; i < ((int)arr.size()); ++i) {
    cin >> arr[i];
    q.push(arr[i]);
    vis.insert(arr[i]);
  }
  sort(arr.begin(), arr.end());
  vector<long long> res = bfs(q, vis, m);
  int ind = 0;
  long long tot = 0;
  sort(res.begin(), res.end());
  for (int i = 0; i < ((int)res.size()); ++i) {
    while (res[i] > arr[ind] && ind != (int)arr.size() - 1) ind++;
    long long adding = 0;
    if (ind == 0)
      adding = abs(arr[ind] - res[i]);
    else
      adding = min(abs(arr[ind] - res[i]), abs(arr[ind - 1] - res[i]));
    tot += adding;
  }
  cout << tot << endl;
  for (int i = 0; i < ((int)res.size()); ++i) cout << res[i] << " ";
  return 0;
}
