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;
long long n, m, t;
long long num[100005];
long long ok = false;
void solve() {
  long long m;
  cin >> n >> m;
  deque<long long> v;
  v.clear();
  v.resize(n, 0);
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  map<long long, bool> M;
  map<long long, long long> Ml;
  map<long long, long long> Mr;
  set<long long> left;
  for (auto k : v) {
    left.insert(k);
    M[k] = 1;
    Ml[k] = k - 1;
    Mr[k] = k + 1;
  }
  long long ctr = 0;
  deque<long long> ans;
  long long sum = 0;
  vector<long long> toer;
  while (m > 0) {
    for (auto k : toer) {
      left.erase(left.find(k));
    }
    toer.clear();
    for (auto k : left) {
      long long flag = 0;
      if (M[Ml[k]]) {
        flag++;
      } else {
        m--;
        ans.push_back(Ml[k]);
        sum += abs(k - Ml[k]);
        M[Ml[k]] = 1;
        Ml[k]--;
        if (m == 0) break;
      }
      if (M[Mr[k]]) {
        flag++;
      } else {
        m--;
        ans.push_back(Mr[k]);
        sum += abs(k - Mr[k]);
        M[Mr[k]] = 1;
        Mr[k]++;
        if (m == 0) break;
      }
      if (flag == 2) {
        toer.push_back(k);
      }
    }
  }
  cout << sum << endl;
  for (auto k : ans) {
    cout << k << " ";
  }
  cout << endl;
}
int32_t main() {
  cin.tie(0);
  ios_base::sync_with_stdio(NULL);
  t = 1;
  while (t--) solve();
  return 0;
}
