#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10;
int n, m;
set<pair<int, int> > s;
map<int, bool> bio;
long long ans;
vector<int> sol;
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    s.insert({1, x - 1});
    s.insert({1, x + 1});
    bio[x] = true;
  }
  while (m > 0) {
    while (true) {
      if (s.size() == 0) break;
      if (bio[s.begin()->second] == true) {
        s.erase(s.begin());
      } else {
        break;
      }
    }
    ans += (long long)s.begin()->first;
    sol.push_back(s.begin()->second);
    bio[s.begin()->second] = true;
    if (!bio[s.begin()->second + 1])
      s.insert({s.begin()->first + 1, s.begin()->second + 1});
    if (!bio[s.begin()->second - 1])
      s.insert({s.begin()->first + 1, s.begin()->second - 1});
    s.erase(s.begin());
    m--;
  }
  printf("%lld\n", ans);
  for (auto it : sol) printf("%d ", it);
  printf("\n");
  return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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 N = 2e5 + 10;
int n, m;
set<pair<int, int> > s;
map<int, bool> bio;
long long ans;
vector<int> sol;
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    s.insert({1, x - 1});
    s.insert({1, x + 1});
    bio[x] = true;
  }
  while (m > 0) {
    while (true) {
      if (s.size() == 0) break;
      if (bio[s.begin()->second] == true) {
        s.erase(s.begin());
      } else {
        break;
      }
    }
    ans += (long long)s.begin()->first;
    sol.push_back(s.begin()->second);
    bio[s.begin()->second] = true;
    if (!bio[s.begin()->second + 1])
      s.insert({s.begin()->first + 1, s.begin()->second + 1});
    if (!bio[s.begin()->second - 1])
      s.insert({s.begin()->first + 1, s.begin()->second - 1});
    s.erase(s.begin());
    m--;
  }
  printf("%lld\n", ans);
  for (auto it : sol) printf("%d ", it);
  printf("\n");
  return 0;
}

```