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 
#include <bits/stdc++.h>
using namespace std;
template <typename T>
void operator<<(ostream& c, vector<T>& a) {
  for (T el : a) {
    c << el << ' ';
  }
  cout << '\n';
}
int main() {
  int n, m;
  cin >> n >> m;
  vector<int> trees(n);
  long long res = 0;
  for (int i = 0; i < n; ++i) {
    cin >> trees[i];
  }
  sort(trees.begin(), trees.end());
  set<int> left, right;
  for (int i = 0; i < n; ++i) {
    left.insert(i);
    right.insert(i);
  }
  vector<int> ans;
  for (int r = 1; m != 0; ++r) {
    auto it = left.begin();
    for (; it != left.end();) {
      int tree = *it;
      if (tree == 0 || trees[tree] - trees[tree - 1] - 1 >= 2 * r) {
        res += r;
        ans.push_back(trees[tree] - r);
        --m;
        if (m == 0) {
          break;
        }
        ++it;
      } else {
        auto it2 = it;
        ++it;
        left.erase(it2);
      }
    }
    if (m == 0) {
      break;
    }
    it = right.begin();
    for (; it != right.end();) {
      int tree = *it;
      if (tree == n - 1 || trees[tree + 1] - trees[tree] - 1 >= 2 * r - 1) {
        res += r;
        ans.push_back(trees[tree] + r);
        --m;
        if (m == 0) {
          break;
        }
        ++it;
      } else {
        auto it2 = it;
        ++it;
        right.erase(it2);
      }
    }
  }
  cout << res << '\n' << ans;
  return 0;
}
