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 int MOD = 1000000007;
const long long INF = 1e18;
const long double PI = 4 * atan((long double)1);
int main() {
  int n, m;
  cin >> n >> m;
  vector<int> trees;
  vector<pair<int, int> > bounds;
  for (int i = 0; i < (n); i++) {
    int x;
    cin >> x;
    trees.push_back(x);
  }
  sort(begin(trees), end(trees));
  priority_queue<pair<int, pair<int, pair<int, int> > >,
                 vector<pair<int, pair<int, pair<int, int> > > >,
                 greater<pair<int, pair<int, pair<int, int> > > > >
      pq;
  for (int i = 0; i < (n); i++) {
    int l, r;
    if (i == 0) {
      l = 1e9;
    } else {
      l = (trees[i] - trees[i - 1] - 1) / 2;
    }
    if (i == n - 1) {
      r = 1e9;
    } else {
      r = (trees[i + 1] - trees[i]) / 2;
    }
    pq.push(make_pair(0, make_pair(i, make_pair(-1, l))));
    pq.push(make_pair(0, make_pair(i, make_pair(1, r))));
  }
  long long overall = 0LL;
  vector<int> ans;
  for (int i = 0; i < (m); i++) {
    while (pq.top().first >= pq.top().second.second.second) {
      pq.pop();
    }
    auto curr = pq.top();
    pq.pop();
    pq.push(make_pair(curr.first + 1, curr.second));
    overall += curr.first * 1LL + 1LL;
    ans.push_back((curr.first + 1) * curr.second.second.first +
                  trees[curr.second.first]);
  }
  cout << overall << '\n';
  for (int i = 0; i < (m); i++) {
    cout << ans[i] << ' ';
  }
}
