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 int N = 2E5 + 7;
const int dir[] = {1, -1};
map<int, bool> visited;
deque<pair<int, int> > Q;
long long sum;
int m, n;
int a[N];
vector<int> res;
void BFS() {
  while (!Q.empty()) {
    int tmp = Q.front().first;
    int weight = Q.front().second;
    Q.pop_front();
    for (int i = 0; i < 2; ++i) {
      if (i == 0) {
        int tmp2 = tmp + dir[i];
        if (visited[tmp2] == false && m > 0) {
          visited[tmp2] = true;
          Q.push_back(pair<int, int>(tmp2, weight + 1));
          res.push_back(tmp2);
          sum = sum + Q.back().second;
          m--;
        }
      }
      if (i == 1) {
        int tmp2 = tmp + dir[i];
        if (visited[tmp2] == false && m > 0) {
          visited[tmp2] = true;
          Q.push_back(pair<int, int>(tmp2, weight + 1));
          res.push_back(tmp2);
          sum = sum + Q.back().second;
          m--;
        }
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(NULL);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    visited[a[i]] = true;
    Q.push_back(pair<int, int>(a[i], 0));
  }
  BFS();
  cout << sum << "\n";
  for (int i = 0; i < res.size(); ++i) cout << res[i] << " ";
}
