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 nax = 1e5 + 5;
vector<long long> adj[nax];
map<long long, long long> visited;
long long td = 0;
vector<long long> v;
long long cnt;
signed main() {
  long long n, m;
  queue<pair<long long, long long>> q;
  cin >> n >> m;
  while (n--) {
    long long a;
    cin >> a;
    q.push({a, 0});
    visited[a] = 1;
  }
  while (!q.empty()) {
    long long cur_pt = q.front().first;
    long long dis = q.front().second;
    q.pop();
    if (v.size() <= m - 1) {
      if (dis != 0) {
        td += dis;
        v.push_back(cur_pt);
      }
    } else {
      break;
    }
    if (!visited.count(cur_pt - 1)) {
      visited[cur_pt - 1] = 1;
      q.push({cur_pt - 1, dis + 1});
    }
    if (!visited.count(cur_pt + 1)) {
      visited[cur_pt + 1] = 1;
      q.push({cur_pt + 1, dis + 1});
    }
  }
  cout << td << endl;
  for (auto u : v) {
    cout << u << " ";
  }
  cout << endl;
  return 0;
}
