#include <bits/stdc++.h>
using namespace std;
void Onigiri() {
  int n, m;
  scanf("%d%d", &n, &m);
  int arr[n];
  priority_queue<pair<int, int>, std::vector<pair<int, int>>,
                 greater<pair<int, int>>>
      l, r;
  set<int> humans, trees;
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    trees.insert(arr[i]);
  }
  sort(arr, arr + n);
  for (int i = 0; i < n; i++) {
    l.push({1, arr[i]});
    r.push({1, arr[i]});
  }
  long long cnt = 0, t = 1000;
  ;
  int inf = 200005;
  while (m != 0) {
    int ll = inf, rr = inf, pr, pl;
    pair<int, int> ltop = l.top();
    pair<int, int> rtop = r.top();
    ll = (ltop).first;
    pl = ltop.second;
    rr = (rtop).first;
    pr = rtop.second;
    if (trees.find(pr + rr) != trees.end() ||
        humans.find(pr + rr) != humans.end()) {
      r.pop();
      rr = inf;
    }
    if (trees.find(pl - ll) != trees.end() ||
        humans.find(pl - ll) != humans.end()) {
      l.pop();
      ll = inf;
    }
    if (l.size() == 0) break;
    if (r.size() == 0) break;
    if (rr == inf || ll == inf) continue;
    if (rr <= ll) {
      cnt += rr;
      m--;
      humans.insert(pr + rr);
      pair<int, int> p = rtop;
      p.first += 1;
      r.pop();
      if (trees.find(pr + rr + 1) == trees.end() ||
          humans.find(pr + rr + 1) == humans.end())
        r.push(p);
    } else {
      cnt += ll;
      m--;
      humans.insert(pl - ll);
      pair<int, int> p = ltop;
      p.first += 1;
      l.pop();
      if (trees.find(pl - ll - 1) == trees.end() ||
          humans.find(pl - ll - 1) == humans.end())
        l.push(p);
    }
  }
  printf("%lld\n", cnt);
  for (auto x : humans) {
    printf("%d ", x);
  }
}
signed main() {
  int t = 1;
  while (t--) {
    Onigiri();
  }
  return 0;
}

### Prompt

Please formulate 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;
void Onigiri() {
  int n, m;
  scanf("%d%d", &n, &m);
  int arr[n];
  priority_queue<pair<int, int>, std::vector<pair<int, int>>,
                 greater<pair<int, int>>>
      l, r;
  set<int> humans, trees;
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    trees.insert(arr[i]);
  }
  sort(arr, arr + n);
  for (int i = 0; i < n; i++) {
    l.push({1, arr[i]});
    r.push({1, arr[i]});
  }
  long long cnt = 0, t = 1000;
  ;
  int inf = 200005;
  while (m != 0) {
    int ll = inf, rr = inf, pr, pl;
    pair<int, int> ltop = l.top();
    pair<int, int> rtop = r.top();
    ll = (ltop).first;
    pl = ltop.second;
    rr = (rtop).first;
    pr = rtop.second;
    if (trees.find(pr + rr) != trees.end() ||
        humans.find(pr + rr) != humans.end()) {
      r.pop();
      rr = inf;
    }
    if (trees.find(pl - ll) != trees.end() ||
        humans.find(pl - ll) != humans.end()) {
      l.pop();
      ll = inf;
    }
    if (l.size() == 0) break;
    if (r.size() == 0) break;
    if (rr == inf || ll == inf) continue;
    if (rr <= ll) {
      cnt += rr;
      m--;
      humans.insert(pr + rr);
      pair<int, int> p = rtop;
      p.first += 1;
      r.pop();
      if (trees.find(pr + rr + 1) == trees.end() ||
          humans.find(pr + rr + 1) == humans.end())
        r.push(p);
    } else {
      cnt += ll;
      m--;
      humans.insert(pl - ll);
      pair<int, int> p = ltop;
      p.first += 1;
      l.pop();
      if (trees.find(pl - ll - 1) == trees.end() ||
          humans.find(pl - ll - 1) == humans.end())
        l.push(p);
    }
  }
  printf("%lld\n", cnt);
  for (auto x : humans) {
    printf("%d ", x);
  }
}
signed main() {
  int t = 1;
  while (t--) {
    Onigiri();
  }
  return 0;
}

```