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;
signed main() {
  long long n, m;
  cin >> n >> m;
  vector<long long> v(n);
  map<long long, long long> mp;
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    mp[v[i]]++;
  }
  set<pair<long long, long long>> s;
  for (long long i = 0; i < n; i++) {
    if (mp[v[i] - 1] == 0) {
      mp[v[i] - 1] = 1;
      s.insert({1, v[i] - 1});
    }
    if (mp[v[i] + 1] == 0) {
      mp[v[i] + 1] = 1;
      s.insert({1, v[i] + 1});
    }
  }
  sort(v.begin(), v.end());
  vector<long long> ansv;
  long long ans = 0;
  for (long long i = 0; i < m; i++) {
    pair<long long, long long> a = *s.begin();
    s.erase(a);
    ans += a.first;
    ansv.push_back(a.second);
    if (mp[a.second - 1] == 0) {
      long long l = 0, r = n + 1;
      mp[a.second - 1] = 1;
      while (l + 1 < r) {
        long long m1 = (l + r) / 2;
        if (v[m1] < a.second - 1) {
          l = m1;
        } else {
          r = m1;
        }
      }
      if (l >= n - 1) {
        s.insert({abs(v[n - 1] - a.second + 1), a.second - 1});
      }
      s.insert({min(abs(v[l] - a.second + 1), abs(v[r] - a.second + 1)),
                a.second - 1});
    }
    if (mp[a.second + 1] == 0) {
      long long l = 0, r = n + 1;
      while (l + 1 < r) {
        long long m1 = (l + r) / 2;
        if (v[m1] < a.second + 1) {
          l = m1;
        } else {
          r = m1;
        }
      }
      if (l >= n - 1) {
        s.insert({abs(v[n - 1] - a.second - 1), a.second + 1});
      }
      s.insert({min(abs(v[l] - a.second - 1), abs(v[r] - a.second - 1)),
                a.second + 1});
      mp[a.second + 1] = 1;
    }
  }
  cout << ans << endl;
  for (long long i = 0; i < ansv.size(); i++) {
    cout << ansv[i] << " ";
  }
}
