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>
#pragma GCC optimize("O2")
using namespace std;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> t(n);
  map<int, bool> vis;
  vector<int> ans;
  long long sum = 0;
  for (int i = 0; i < n; i++) {
    cin >> t[i];
    vis[t[i]] = 1;
  }
  priority_queue<pair<int, int>, vector<pair<int, int>>,
                 greater<pair<int, int>>>
      q;
  for (int i = 0; i < n; i++) {
    q.push({1, t[i] + 1}), q.push({1, t[i] - 1});
  }
  sort(t.begin(), t.end());
  while (m) {
    auto x = q.top();
    q.pop();
    if (vis[x.second]) continue;
    m--;
    vis[x.second] = 1;
    sum += x.first;
    ans.push_back(x.second);
    int cur = x.second;
    for (auto nx : {cur - 1, cur + 1}) {
      if (vis[nx]) continue;
      auto it = lower_bound(t.begin(), t.end(), nx);
      int mn = 2e9 + 1;
      if (it == t.end()) {
        mn = min(mn, nx - t.back());
      } else {
        mn = min(mn, abs(nx - *it));
        if (it != t.begin()) {
          it--;
          mn = min(mn, abs(nx - *it));
        }
      }
      q.push({mn, nx});
    }
  }
  cout << sum << "\n";
  for (auto x : ans) cout << x << " ";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
}
