#include <bits/stdc++.h>
using namespace std;
const long double PI = acos(-1);
long long powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
map<long long, bool> vis;
long long x[200005], y[200005];
set<pair<long long, long long> > s;
void solve() {
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> x[i];
    vis[x[i]] = true;
    s.insert({0, x[i]});
  }
  long long ans = 0;
  long long i = 1;
  while (i <= m) {
    pair<long long, long long> t = *s.begin();
    s.erase(t);
    if (!vis[t.second]) {
      ans += t.first;
      y[i] = t.second;
      i++;
    }
    if (!vis[t.second - 1]) s.insert({t.first + 1, t.second - 1});
    if (!vis[t.second + 1]) s.insert({t.first + 1, t.second + 1});
    vis[t.second] = 1;
  }
  cout << ans << endl;
  for (long long k = 1; k <= m; k++) cout << y[k] << " ";
  cout << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long double PI = acos(-1);
long long powm(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return res;
}
map<long long, bool> vis;
long long x[200005], y[200005];
set<pair<long long, long long> > s;
void solve() {
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) {
    cin >> x[i];
    vis[x[i]] = true;
    s.insert({0, x[i]});
  }
  long long ans = 0;
  long long i = 1;
  while (i <= m) {
    pair<long long, long long> t = *s.begin();
    s.erase(t);
    if (!vis[t.second]) {
      ans += t.first;
      y[i] = t.second;
      i++;
    }
    if (!vis[t.second - 1]) s.insert({t.first + 1, t.second - 1});
    if (!vis[t.second + 1]) s.insert({t.first + 1, t.second + 1});
    vis[t.second] = 1;
  }
  cout << ans << endl;
  for (long long k = 1; k <= m; k++) cout << y[k] << " ";
  cout << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  for (long long i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```