#include <bits/stdc++.h>
using namespace std;
const long long N = 1e6 + 10;
queue<pair<long long, long long>> que;
map<long long, bool> st;
long long ans, a[N], cnt;
long long t = 0, n, m;
void solve() {
  while (que.size()) {
    auto t = que.front();
    que.pop();
    if (st[t.second]) continue;
    st[t.second] = true, a[cnt++] = t.second, ans += t.first;
    if (!st[t.second - 1]) que.push({t.first + 1, t.second - 1});
    if (!st[t.second + 1]) que.push({t.first + 1, t.second + 1});
    if (cnt == m) break;
  }
}
signed main() {
  for (cin >> n >> m; n--; cin >> a[t++], st[a[t - 1]] = true)
    ;
  for (long long(i) = 0; (i) < (t); (i)++)
    que.push({1, a[i] - 1}), que.push({1, a[i] + 1});
  solve();
  cout << ans << '\n';
  for (long long(i) = 0; (i) < (cnt); (i)++) cout << a[i] << ' ';
  return 0;
}

### Prompt

Please formulate a cpp solution to the following problem:

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 long N = 1e6 + 10;
queue<pair<long long, long long>> que;
map<long long, bool> st;
long long ans, a[N], cnt;
long long t = 0, n, m;
void solve() {
  while (que.size()) {
    auto t = que.front();
    que.pop();
    if (st[t.second]) continue;
    st[t.second] = true, a[cnt++] = t.second, ans += t.first;
    if (!st[t.second - 1]) que.push({t.first + 1, t.second - 1});
    if (!st[t.second + 1]) que.push({t.first + 1, t.second + 1});
    if (cnt == m) break;
  }
}
signed main() {
  for (cin >> n >> m; n--; cin >> a[t++], st[a[t - 1]] = true)
    ;
  for (long long(i) = 0; (i) < (t); (i)++)
    que.push({1, a[i] - 1}), que.push({1, a[i] + 1});
  solve();
  cout << ans << '\n';
  for (long long(i) = 0; (i) < (cnt); (i)++) cout << a[i] << ' ';
  return 0;
}

```