#include <bits/stdc++.h>
using namespace std;
template <typename T>
using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
const int INF = 1 << 29;
const long long INFL = 1LL << 60;
const long long mod = 1000000007;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  long long dst = 0;
  map<int, int> mm;
  queue<pair<int, int>> q;
  for (int i = 0; i < int(n); i++) {
    int a;
    cin >> a;
    mm[a] = 1;
    q.push({a, 1});
  }
  vector<int> ans;
  while (true) {
    while (!q.empty()) {
      int a = q.front().first;
      int cur = q.front().second;
      q.pop();
      if (mm[a + 1] == 0) {
        mm[a + 1] = 1;
        ans.push_back(a + 1);
        m--;
        dst += cur;
        q.push({a + 1, cur + 1});
        if (m == 0) {
          cout << dst << endl;
          for (int j : ans) {
            cout << j << ' ';
          }
          return 0;
        }
      }
      if (mm[a - 1] == 0) {
        mm[a - 1] = 1;
        ans.push_back(a - 1);
        m--;
        dst += cur;
        q.push({a - 1, cur + 1});
        if (m == 0) {
          cout << dst << endl;
          for (int j : ans) {
            cout << j << ' ';
          }
          return 0;
        }
      }
    }
  }
  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;
template <typename T>
using min_priority_queue = priority_queue<T, vector<T>, greater<T>>;
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
const int INF = 1 << 29;
const long long INFL = 1LL << 60;
const long long mod = 1000000007;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  long long dst = 0;
  map<int, int> mm;
  queue<pair<int, int>> q;
  for (int i = 0; i < int(n); i++) {
    int a;
    cin >> a;
    mm[a] = 1;
    q.push({a, 1});
  }
  vector<int> ans;
  while (true) {
    while (!q.empty()) {
      int a = q.front().first;
      int cur = q.front().second;
      q.pop();
      if (mm[a + 1] == 0) {
        mm[a + 1] = 1;
        ans.push_back(a + 1);
        m--;
        dst += cur;
        q.push({a + 1, cur + 1});
        if (m == 0) {
          cout << dst << endl;
          for (int j : ans) {
            cout << j << ' ';
          }
          return 0;
        }
      }
      if (mm[a - 1] == 0) {
        mm[a - 1] = 1;
        ans.push_back(a - 1);
        m--;
        dst += cur;
        q.push({a - 1, cur + 1});
        if (m == 0) {
          cout << dst << endl;
          for (int j : ans) {
            cout << j << ' ';
          }
          return 0;
        }
      }
    }
  }
  return 0;
}

```