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;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  queue<pair<pair<int, int>, int> > vec;
  set<int> done;
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    vec.push({{a - 1, a}, -1});
    vec.push({{a + 1, a}, 1});
    done.insert(a);
  }
  long long ans = 0;
  vector<int> ff;
  done.insert(2e9 + 5);
  while (m) {
    pair<pair<int, int>, int> pp = vec.front();
    vec.pop();
    if (done.find(pp.first.first) != done.end()) {
      pp.first.first == 2e9 + 5;
    } else {
      ans += abs(pp.first.first - pp.first.second);
      ff.push_back(pp.first.first);
      m--;
      done.insert(pp.first.first);
      pp.first.first += pp.second;
      vec.push(pp);
      if (m == 0) break;
    }
  }
  printf("%lld\n", ans);
  for (int i = 0; i < ff.size(); i++) {
    printf("%d ", ff[i]);
  }
  return 0;
}
