#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
using uint = unsigned int;
using ull = unsigned long long;
template <typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
using ull = unsigned long long;
using vi = vector<int>;
using vii = vector<pii>;
using si = set<int>;
using msi = multiset<int>;
using ii = pair<int, int>;
using iii = pair<int, ii>;
ll ans;
vector<int> people;
void dijkstra(int pos[], int n, int m) {
  queue<int> Q;
  map<int, int> dis;
  for (int i = 0; i < n; i++) {
    dis[pos[i]] = 0;
    Q.push(pos[i]);
  }
  while (!Q.empty()) {
    if (people.size() == m) return;
    int newq = Q.front();
    Q.pop();
    if (dis[newq] != 0) {
      ans += dis[newq];
      people.push_back(newq);
    }
    if (!dis.count(newq - 1)) {
      dis[newq - 1] = dis[newq] + 1;
      Q.push(newq - 1);
    }
    if (!dis.count(newq + 1)) {
      dis[newq + 1] = dis[newq] + 1;
      Q.push(newq + 1);
    }
  }
}
int main() {
  int n, m;
  cin >> n >> m;
  int pos[n];
  for (int i = 0; i < n; i++) cin >> pos[i];
  dijkstra(pos, n, m);
  cout << ans << endl;
  for (int i = 0; i < m; i++) {
    cout << people[i] << " ";
  }
}

### 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;
using ll = long long;
using ld = long double;
using uint = unsigned int;
using ull = unsigned long long;
template <typename T>
using pair2 = pair<T, T>;
using pii = pair<int, int>;
using pli = pair<ll, int>;
using pll = pair<ll, ll>;
using ull = unsigned long long;
using vi = vector<int>;
using vii = vector<pii>;
using si = set<int>;
using msi = multiset<int>;
using ii = pair<int, int>;
using iii = pair<int, ii>;
ll ans;
vector<int> people;
void dijkstra(int pos[], int n, int m) {
  queue<int> Q;
  map<int, int> dis;
  for (int i = 0; i < n; i++) {
    dis[pos[i]] = 0;
    Q.push(pos[i]);
  }
  while (!Q.empty()) {
    if (people.size() == m) return;
    int newq = Q.front();
    Q.pop();
    if (dis[newq] != 0) {
      ans += dis[newq];
      people.push_back(newq);
    }
    if (!dis.count(newq - 1)) {
      dis[newq - 1] = dis[newq] + 1;
      Q.push(newq - 1);
    }
    if (!dis.count(newq + 1)) {
      dis[newq + 1] = dis[newq] + 1;
      Q.push(newq + 1);
    }
  }
}
int main() {
  int n, m;
  cin >> n >> m;
  int pos[n];
  for (int i = 0; i < n; i++) cin >> pos[i];
  dijkstra(pos, n, m);
  cout << ans << endl;
  for (int i = 0; i < m; i++) {
    cout << people[i] << " ";
  }
}

```