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 long long maxn = 2e5 + 7;
long long n, m, x[maxn], e[maxn], ans = 0, p = 0;
map<long long, long long> vis;
vector<long long> res;
void solve() {
  cin >> n >> m;
  for (long long i = 0; i < n; ++i) {
    cin >> x[i];
    vis[x[i]] = 1;
    e[i] = i + 1;
  }
  sort(x, x + n);
  for (long long i = 1; m > 0; ++i) {
    for (long long j = 0; j < n; j = e[j]) {
      if (j < n - 1 && vis[x[j] - i] && x[j] + i >= x[j + 1] - i) {
        e[p] = e[j];
      }
      if (!vis[x[j] - i]) {
        vis[x[j] - i] = 2;
        res.push_back(x[j] - i);
        ans += i, m--;
      }
      if (m <= 0) {
        break;
      }
      if (!vis[x[j] + i]) {
        vis[x[j] + i] = 2;
        res.push_back(x[j] + i);
        ans += i, m--;
      }
      p = j;
      if (m <= 0) {
        break;
      }
    }
  }
  cout << ans << '\n';
  for (long long i = 0; i < res.size(); ++i) {
    cout << res[i];
    if (i < (long long)res.size() - 1) {
      cout << ' ';
    }
  }
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
