#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
int qnt[200200];
int x[200200];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  map<int, int> mapa;
  for (int i = 0; i < n; i++) {
    cin >> x[i];
    mapa[x[i]] = 1;
  }
  set<pair<int, int> > pec;
  for (int i = 0; i < n; i++) {
    int ant = x[i] - 1;
    int prox = x[i] + 1;
    if (mapa[ant] == 0) pec.insert({1, ant});
    if (mapa[prox] == 0) pec.insert({1, prox});
  }
  int cnt = 0;
  long long int distancia = 0;
  vector<int> ans;
  while (cnt < m) {
    int topo = (*pec.begin()).second;
    int custo = (*pec.begin()).first;
    pec.erase(*pec.begin());
    int ant = topo - 1;
    int prox = topo + 1;
    if (mapa[topo] == 0) {
      mapa[topo] = 1;
      distancia += custo;
      ans.push_back(topo);
      cnt++;
      if (mapa[ant] == 0) pec.insert({custo + 1, ant});
      if (mapa[prox] == 0) pec.insert({custo + 1, prox});
    }
  }
  cout << distancia << endl;
  for (auto y : ans) {
    cout << y << " ";
  }
  cout << endl;
}

### Prompt

Please provide a cpp coded solution 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;
const int INF = 0x3f3f3f3f;
int qnt[200200];
int x[200200];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  map<int, int> mapa;
  for (int i = 0; i < n; i++) {
    cin >> x[i];
    mapa[x[i]] = 1;
  }
  set<pair<int, int> > pec;
  for (int i = 0; i < n; i++) {
    int ant = x[i] - 1;
    int prox = x[i] + 1;
    if (mapa[ant] == 0) pec.insert({1, ant});
    if (mapa[prox] == 0) pec.insert({1, prox});
  }
  int cnt = 0;
  long long int distancia = 0;
  vector<int> ans;
  while (cnt < m) {
    int topo = (*pec.begin()).second;
    int custo = (*pec.begin()).first;
    pec.erase(*pec.begin());
    int ant = topo - 1;
    int prox = topo + 1;
    if (mapa[topo] == 0) {
      mapa[topo] = 1;
      distancia += custo;
      ans.push_back(topo);
      cnt++;
      if (mapa[ant] == 0) pec.insert({custo + 1, ant});
      if (mapa[prox] == 0) pec.insert({custo + 1, prox});
    }
  }
  cout << distancia << endl;
  for (auto y : ans) {
    cout << y << " ";
  }
  cout << endl;
}

```