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 fastik() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
}
void solve() {}
long long getDist(long long a, long long b) { return abs(a - b); }
int main() {
  fastik();
  long long n, m;
  cin >> n >> m;
  vector<long long> x(n);
  map<long long, long long> used;
  for (int i = 0; i < (int)(n); ++i) {
    cin >> x[i];
    used[x[i]] = 2;
  }
  vector<long long> l(n);
  vector<long long> r(n);
  vector<long long> dl(n, 0);
  vector<long long> dr(n, 0);
  vector<long long> ans;
  sort((x).begin(), (x).end());
  for (int i = 0; i < (int)(n); ++i) {
    l[i] = x[i] - 1;
    r[i] = x[i] + 1;
  }
  long long dd = getDist(x[0], x[n - 1]) - (n - 2) - 1;
  long long k = m;
  long long t = 0;
  long long f = -1;
  long long res = 0;
  set<long long> ind;
  for (int i = 0; i < (int)(n); ++i) {
    ind.insert(i);
  }
  while (k) {
    t++;
    if (t % 2 == 1) {
      for (auto i : ind) {
        if (used[l[i]]) {
          dl[i] = 1;
        }
        if (dl[i]) {
          continue;
        }
        used[l[i]]++;
        ans.push_back(l[i]);
        res += getDist(l[i], x[i]);
        k--;
        if (k == 0) {
          f = 1;
          break;
        }
        l[i]--;
      }
    } else {
      for (auto i : ind) {
        if (used[r[i]]) {
          dr[i] = 1;
        }
        if (dr[i]) continue;
        used[r[i]]++;
        ans.push_back(r[i]);
        res += getDist(r[i], x[i]);
        k--;
        if (k == 0) {
          f = 1;
          break;
        }
        r[i]++;
      }
    }
    vector<long long> del;
    for (auto i : ind) {
      if (dl[i] && dr[i]) {
        del.push_back(i);
      }
    }
    for (auto j : del) {
      ind.erase(j);
    }
    if (f == 1) {
      break;
    }
  }
  cout << res << "\n";
  for (auto i : ans) cout << i << " ";
}
