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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  long long ans = 0;
  long long arr[n];
  set<pair<int, int> > s;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  sort(arr, arr + n);
  s.insert({INT_MIN, arr[0]});
  for (int i = 1; i < n; i++) {
    s.insert({arr[i - 1], arr[i]});
  }
  s.insert({arr[n - 1], INT_MAX});
  int cnt = 0;
  set<pair<int, int> >::iterator it = s.begin();
  vector<int> res;
  map<pair<int, int>, int> mp;
  while (cnt < k) {
    pair<int, int> x;
    set<pair<int, int> >::iterator it1;
    if ((*it).first == INT_MIN) {
      res.push_back((*it).second - 1);
      cnt += 1;
      x.first = (*it).first;
      x.second = (*it).second - 1;
      mp[x] = mp[(*it)] + 1;
      ans += mp[x];
      it1 = it;
      it1++;
      s.erase(it);
      s.insert(x);
    } else if ((*it).second == INT_MAX) {
      res.push_back((*it).first + 1);
      cnt += 1;
      x.first = (*it).first + 1;
      x.second = (*it).second;
      mp[x] = mp[(*it)] + 1;
      ans += mp[x];
      it1 = it;
      it1++;
      s.erase(it);
      s.insert(x);
    } else {
      if ((*it).first + 1 == (*it).second) {
        it1 = it;
        it1++;
        s.erase(it);
      } else {
        res.push_back((*it).second - 1);
        cnt += 1;
        mp[{(*it).first, (*it).second - 1}] = 1 + mp[*it];
        ans += mp[{(*it).first, (*it).second - 1}];
        if (cnt == k) break;
        if ((*it).first + 1 == (*it).second - 1) {
          it1 = it;
          it1++;
          s.erase(it);
        } else {
          res.push_back((*it).first + 1);
          cnt += 1;
          x.first = (*it).first + 1;
          x.second = (*it).second - 1;
          mp[x] = mp[(*it)] + 1;
          ans += mp[x];
          it1 = it;
          it1++;
          s.erase(it);
          s.insert(x);
        }
      }
    }
    if (it1 == s.end())
      it = s.begin();
    else
      it = it1;
  }
  cout << ans << endl;
  for (int i = 0; i < res.size(); i++) cout << res[i] << " ";
  cout << endl;
  return 0;
}
