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;
long long mod = 1e9 + 7;
long long cl(long double a) {
  if (a > (long long)a) {
    return (long long)a + 1;
  } else {
    return (long long)a;
  }
}
long long flr(long double a) { return (long long)a; }
const long long M = 2e5 + 100;
long long n, m, a[M];
void solve() {
  cin >> n;
  cin >> m;
  for (long long i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  map<long long, long long> mark;
  map<long long, long long> dist;
  vector<long long> res;
  long long ans = 0;
  queue<long long> q;
  for (long long i = 0; i < n; i++) {
    q.push(a[i]);
    mark[a[i]] = 1;
  }
  while (!q.empty()) {
    long long par = q.front();
    q.pop();
    if (!mark[par - 1]) {
      res.push_back(par - 1);
      dist[par - 1] = dist[par] + 1;
      q.push(par - 1);
      ans += dist[par - 1];
      mark[par - 1] = 1;
    }
    if (res.size() == m) break;
    if (!mark[par + 1]) {
      res.push_back(par + 1);
      dist[par + 1] = dist[par] + 1;
      q.push(par + 1);
      ans += dist[par + 1];
      mark[par + 1] = 1;
    }
    if (res.size() == m) break;
  }
  cout << ans << "\n";
  for (long long i = 0; i < m; i++) cout << res[i] << " ";
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  while (t--) solve();
}
