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;
bool compare(pair<long long, pair<long long, long long> >& a,
             pair<long long, pair<long long, long long> >& b) {
  return (a.second.second - a.second.first) >
         (b.second.second - b.second.first);
}
int main() {
  long long n, m;
  cin >> n >> m;
  vector<long long> a(n);
  vector<pair<long long, pair<long long, long long> > > empty;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  for (int i = 0; i < n; i++) {
    if (i == 0)
      empty.push_back(make_pair(1, make_pair(INT_MIN, a[i] - 1)));
    else if (a[i - 1] + 1 <= a[i] - 1)
      empty.push_back(make_pair(1, make_pair(a[i - 1] + 1, a[i] - 1)));
  }
  empty.push_back(make_pair(1, make_pair(a[n - 1] + 1, INT_MAX)));
  sort(empty.begin(), empty.end(), compare);
  int i = 0;
  vector<int> ans;
  long long dist = 0;
  int end = empty.size() - 1;
  while (m > 0) {
    if (i > end) i = 0;
    if (empty[i].second.first == INT_MIN) {
      dist += empty[i].first;
      ans.push_back(empty[i].second.second);
      empty[i] = make_pair(
          empty[i].first + 1,
          make_pair(empty[i].second.first, empty[i].second.second - 1));
      m--;
    } else if (empty[i].second.second == INT_MAX) {
      dist += empty[i].first;
      ans.push_back(empty[i].second.first);
      empty[i] = make_pair(
          empty[i].first + 1,
          make_pair(empty[i].second.first + 1, empty[i].second.second));
      m--;
    } else if (empty[i].second.second == empty[i].second.first) {
      dist += empty[i].first;
      ans.push_back(empty[i].second.second);
      empty[i] = make_pair(-1, make_pair(-1, -2));
      m--;
    } else if (empty[i].second.first > empty[i].second.second) {
      i = 0;
      continue;
    } else if (empty[i].second.first < empty[i].second.second) {
      long long sff = empty[i].second.first;
      long long sss = empty[i].second.second;
      long long fff = empty[i].first;
      ans.push_back(sff);
      sff++;
      dist += fff;
      if (m == 1) break;
      ans.push_back(sss);
      sss--;
      dist += fff;
      if (sff <= sss)
        fff += 1;
      else
        fff = -1;
      empty[i] = make_pair(fff, make_pair(sff, sss));
      m -= 2;
    }
    i++;
  }
  cout << dist << endl;
  for (int x : ans) cout << x << " ";
  cout << endl;
  return 0;
}
