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;
void fastInOut();
const int d[] = {-1, 1};
int n, m, x;
priority_queue<pair<int, int> > Q;
set<int> vis;
vector<int> ret;
long long solve() {
  long long tot = 0;
  while (m) {
    pair<int, int> cur = Q.top();
    Q.pop();
    if (vis.find(cur.second) != vis.end()) continue;
    --m, vis.insert(cur.second);
    tot += -cur.first, ret.push_back(cur.second);
    for (int i = 0; i < 2; ++i)
      if (vis.find(cur.second + d[i]) == vis.end())
        Q.push({cur.first - 1, cur.second + d[i]});
  }
  return tot;
}
int main() {
  fastInOut();
  cin >> n >> m;
  while (n--) {
    cin >> x, vis.insert(x);
    Q.push({-1, x - 1}), Q.push({-1, x + 1});
  }
  cout << solve() << "\n";
  sort(ret.begin(), ret.end());
  for (int i = 0; i < int(ret.size()); ++i)
    cout << ret[i] << " \n"[i == int(ret.size()) - 1];
  return 0;
}
void fastInOut() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL), cout.tie(NULL);
}
