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;
struct people {
  int coor = 0;
  int dis = 0;
};
queue<people> q;
map<people, bool> mp;
int n, m, a[200010];
long long total = 0;
vector<int> ans;
bool operator<(people a, people b) { return a.coor < b.coor; }
people ms(int coor, int dis) {
  people temp;
  temp.coor = coor;
  temp.dis = dis;
  return temp;
}
int main() {
  cin >> n >> m;
  for (int(i) = (0); (i) <= (n - 1); ++(i)) {
    int rd;
    cin >> rd;
    people temp = ms(rd, 0);
    q.push(temp);
    mp[temp] = 1;
  }
  while ((!q.empty()) && (ans.size() < m)) {
    people head = q.front();
    q.pop();
    if (head.dis != 0) {
      total += head.dis;
      ans.push_back(head.coor);
    }
    people toLeft = ms(head.coor - 1, head.dis + 1);
    people toRight = ms(head.coor + 1, head.dis + 1);
    if (mp[toLeft] != 1) {
      mp[toLeft] = 1;
      q.push(toLeft);
    }
    if (mp[toRight] != 1) {
      mp[toRight] = 1;
      q.push(toRight);
    }
  }
  cout << total << endl;
  for (int(i) = (0); (i) <= (m - 1); ++(i))
    cout << ans[i] << ((i == m - 1) ? "\n" : " ");
  return 0;
}
