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;
set<int> vis;
vector<int> ans;
map<int, int> d;
int main() {
  int n, m;
  cin >> n >> m;
  queue<int> q;
  int count = 0;
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    q.push(x);
    vis.insert(x);
    d[x] = 0;
  }
  long long int res = 0;
  while (!q.empty()) {
    int now = q.front();
    q.pop();
    if (int(ans.size()) == m) break;
    if (d[now] != 0) {
      res += d[now];
      ans.push_back(now);
    }
    if (vis.find(now - 1) != vis.end())
      res += 0;
    else {
      q.push(now - 1);
      vis.insert(now - 1);
      d[now - 1] = d[now] + 1;
    }
    if (vis.find(now + 1) != vis.end())
      continue;
    else {
      q.push(now + 1);
      vis.insert(now + 1);
      d[now + 1] = d[now] + 1;
    }
  }
  printf("%lld\n", res);
  for (int i = 0; i < ans.size(); i++) {
    printf("%d ", ans[i]);
  }
  cout << endl;
}
