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;
using ld = long double;
template <class T>
inline T gcd(T a, T b) {
  return !a ? b : gcd(b % a, a);
}
template <class T>
inline T lcm(T a, T b) {
  return (a * b) / gcd(a, b);
}
mt19937 rnd(time(0));
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  queue<int> q;
  map<int, int> d;
  for (int i = 0; i < n; i++) {
    d[a[i]] = 0;
    q.push(a[i]);
  }
  long long ans = 0ll;
  vector<int> points;
  while (!q.empty()) {
    if ((int)points.size() == m) {
      break;
    }
    int v = q.front();
    q.pop();
    if (d[v] != 0) {
      ans += d[v];
      points.push_back(v);
    }
    if (!d.count(v + 1)) {
      d[v + 1] = d[v] + 1;
      q.push(v + 1);
    }
    if (!d.count(v - 1)) {
      d[v - 1] = d[v] + 1;
      q.push(v - 1);
    }
  }
  cout << ans << '\n';
  for (int x : points) {
    cout << x << " ";
  }
  return;
}
signed main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int t = 1;
  while (t-- > 0) {
    solve();
  }
  return 0;
}
