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;
using ii = pair<int, int>;
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  int i, n, m;
  cin >> n >> m;
  vector<int> x(n);
  vector<int> rest;
  multimap<int, ii> dist;
  for (i = 0; i < n; i++) cin >> x[i];
  sort(x.begin(), x.end());
  dist.insert({1, {INT_MIN, 0}});
  dist.insert({1, {INT_MAX, n - 1}});
  for (i = 0; i < n - 1; i++) {
    if (x[i + 1] - x[i] >= 2) dist.insert({1, {i, i + 1}});
  }
  i = 0;
  int a, b, valor;
  long long sum = 0;
  auto ini = dist.begin();
  while (i != m) {
    ini = dist.begin();
    a = (ini->second).first;
    b = (ini->second).second;
    valor = ini->first;
    sum += valor;
    if (a == INT_MAX)
      rest.push_back(x[b] + valor);
    else
      rest.push_back(x[b] - valor);
    i++;
    dist.erase(ini);
    if (i != m) {
      if (a != INT_MIN && a != INT_MAX) {
        if (x[b] - x[a] > 2 * valor) {
          i++;
          sum += valor;
          rest.push_back(x[a] + valor);
          if (x[b] - x[a] > 2 * valor + 1) dist.insert({valor + 1, {a, b}});
        }
      } else {
        dist.insert({valor + 1, {a, b}});
      }
    }
  }
  cout << sum << endl;
  for (i = 0; i < rest.size(); i++) cout << rest[i] << ' ';
  cout << endl;
  return 0;
}
