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;
void Onigiri() {
  long long n, m;
  cin >> n >> m;
  long long arr[n];
  priority_queue<pair<long long, long long>,
                 std::vector<pair<long long, long long>>,
                 greater<pair<long long, long long>>>
      l, r;
  set<long long> humans, tree;
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    tree.insert(arr[i]);
  }
  sort(arr, arr + n);
  for (long long i = 0; i < n; i++) {
    l.push({1, arr[i] - 1});
    r.push({1, arr[i] + 1});
  }
  long long cnt = 0, t = 4000005;
  ;
  long long inf = 200005;
  while (m > 0) {
    long long ll = l.top().first;
    long long rr = r.top().first;
    long long lp = l.top().second;
    long long rp = r.top().second;
    if (tree.count(rp) || humans.count(rp)) {
      r.pop();
      continue;
    }
    if (tree.count(lp) || humans.count(lp)) {
      l.pop();
      continue;
    }
    if (rr <= ll) {
      cnt += rr;
      r.pop();
      m--;
      humans.insert(rp);
      r.push({rr + 1, rp + 1});
    } else {
      cnt += ll;
      l.pop();
      m--;
      humans.insert(lp);
      l.push({ll + 1, lp - 1});
    }
  }
  cout << cnt << endl;
  for (auto x : humans) {
    cout << x << " ";
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    Onigiri();
  }
  return 0;
}
