#include <bits/stdc++.h>
using namespace std;
const long long M = 1e9 + 7;
const long long INF = LLONG_MAX / 2;
const long long N = 2e5 + 1;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  {
    long long n, m;
    cin >> n >> m;
    vector<long long> first(n);
    queue<long long> q;
    map<long long, long long> d;
    for (long long i = 0; i < n; i++) {
      cin >> first[i];
      d[first[i]] = 0;
      q.push(first[i]);
    }
    long long ans = 0;
    vector<long long> second;
    while (!q.empty()) {
      if (second.size() == m) break;
      long long a = q.front();
      q.pop();
      if (d[a]) {
        ans += d[a];
        second.push_back(a);
      }
      if (!d.count(a - 1)) {
        d[a - 1] = d[a] + 1;
        q.push(a - 1);
      }
      if (!d.count(a + 1)) {
        d[a + 1] = d[a] + 1;
        q.push(a + 1);
      }
    }
    cout << ans << '\n';
    for (auto i : second) cout << i << " ";
  }
  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 long long M = 1e9 + 7;
const long long INF = LLONG_MAX / 2;
const long long N = 2e5 + 1;
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  {
    long long n, m;
    cin >> n >> m;
    vector<long long> first(n);
    queue<long long> q;
    map<long long, long long> d;
    for (long long i = 0; i < n; i++) {
      cin >> first[i];
      d[first[i]] = 0;
      q.push(first[i]);
    }
    long long ans = 0;
    vector<long long> second;
    while (!q.empty()) {
      if (second.size() == m) break;
      long long a = q.front();
      q.pop();
      if (d[a]) {
        ans += d[a];
        second.push_back(a);
      }
      if (!d.count(a - 1)) {
        d[a - 1] = d[a] + 1;
        q.push(a - 1);
      }
      if (!d.count(a + 1)) {
        d[a + 1] = d[a] + 1;
        q.push(a + 1);
      }
    }
    cout << ans << '\n';
    for (auto i : second) cout << i << " ";
  }
  return 0;
}

```