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 INF = 2000000000000000000LL;
const int inf = 0x3f3f3f3f;
const long double EPS = 1e-9;
int arr[200009];
int bfs() {
  int n, m;
  long long cost = 0;
  cin >> n >> m;
  map<int, int> taken;
  map<pair<int, int>, int> qtaken;
  queue<pair<int, int> > process;
  int value;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    taken[arr[i]] = 0;
  }
  for (int i = 0; i < n; i++) {
    value = arr[i];
    int left = value - 1, right = value + 1;
    if (taken.find(left) == taken.end() && m) {
      taken[left] = 1;
      m--;
      cost++;
      if (qtaken.find({1, value}) == qtaken.end())
        process.push({1, value}), qtaken[{1, value}] = 1;
    }
    if (taken.find(right) == taken.end() && m) {
      taken[right] = 1;
      m--;
      cost++;
      if (qtaken.find({1, value}) == qtaken.end())
        process.push({1, value}), qtaken[{1, value}] = 1;
    }
    if (!m) break;
  }
  pair<int, int> top, now;
  int cst = 0, left, right;
  while (!process.empty() && m) {
    top = process.front();
    process.pop();
    cst = top.first + 1, left = top.second - cst, right = top.second + cst;
    if (taken.find(left) == taken.end() && m) {
      taken[left] = 1;
      m--;
      cost += cst;
      if (qtaken.find({cst, top.second}) == qtaken.end())
        process.push({cst, top.second}), qtaken[{cst, top.second}] = 1;
    }
    if (taken.find(right) == taken.end() && m) {
      taken[right] = 1;
      m--;
      cost += cst;
      if (qtaken.find({cst, top.second}) == qtaken.end())
        process.push({cst, top.second}), qtaken[{cst, top.second}] = 1;
    }
    if (!m) break;
  }
  cout << cost << endl;
  for (auto it : taken)
    if (it.second) cout << it.first << " ";
  return cost;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int cost = bfs();
  return 0;
}
