#include <bits/stdc++.h>
using namespace std;
map<long long, long long> incoada;
queue<pair<long long, long long>> coada;
vector<long long> sol;
int main() {
  long long n, m, x, i, f;
  long long rez = 0;
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    cin >> x;
    incoada[x]++;
    coada.push({x, 0});
  }
  while (m) {
    f = coada.front().first;
    if (f + 1 <= 2e9 && m && !incoada[f + 1]) {
      rez += coada.front().second + 1;
      m--;
      incoada[f + 1]++;
      sol.push_back(f + 1);
      coada.push({f + 1, coada.front().second + 1});
    }
    if (f - 1 >= -2e9 && m && !incoada[f - 1]) {
      rez += coada.front().second + 1;
      m--;
      incoada[f - 1]++;
      sol.push_back(f - 1);
      coada.push({f - 1, coada.front().second + 1});
    }
    coada.pop();
  }
  cout << rez << "\n";
  for (auto x : sol) {
    cout << x << " ";
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
map<long long, long long> incoada;
queue<pair<long long, long long>> coada;
vector<long long> sol;
int main() {
  long long n, m, x, i, f;
  long long rez = 0;
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    cin >> x;
    incoada[x]++;
    coada.push({x, 0});
  }
  while (m) {
    f = coada.front().first;
    if (f + 1 <= 2e9 && m && !incoada[f + 1]) {
      rez += coada.front().second + 1;
      m--;
      incoada[f + 1]++;
      sol.push_back(f + 1);
      coada.push({f + 1, coada.front().second + 1});
    }
    if (f - 1 >= -2e9 && m && !incoada[f - 1]) {
      rez += coada.front().second + 1;
      m--;
      incoada[f - 1]++;
      sol.push_back(f - 1);
      coada.push({f - 1, coada.front().second + 1});
    }
    coada.pop();
  }
  cout << rez << "\n";
  for (auto x : sol) {
    cout << x << " ";
  }
  return 0;
}

```