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;
const int MOD = 1e9 + 7;
const double eps = 1e-10;
const long long INF = 1e18;
const int N = 510;
void solve() {
  int n, m;
  cin >> n >> m;
  queue<int> q;
  map<int, int> dist;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    q.push(x);
    dist[x] = 0;
  }
  int curr = 0;
  while (!q.empty()) {
    int top = q.front();
    q.pop();
    if (dist.find(top + 1) == dist.end()) {
      dist[top + 1] = dist[top] + 1;
      curr++;
      if (curr == m) break;
      q.push(top + 1);
    }
    if (dist.find(top - 1) == dist.end()) {
      dist[top - 1] = dist[top] + 1;
      curr++;
      if (curr == m) break;
      q.push(top - 1);
    }
  }
  long long ans = 0;
  vector<int> res;
  for (auto it : dist) {
    if (it.second) res.push_back(it.first);
    ans += it.second;
  }
  cout << ans << '\n';
  for (auto it : res) cout << it << ' ';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt = 1;
  while (tt--) {
    solve();
  }
  return 0;
}
