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 preprocess(void) { return; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.precision(20);
  preprocess();
  int n, m;
  cin >> n >> m;
  vector<int> x(n);
  for (auto i = (0); i < n; i++) cin >> x[i];
  sort(x.begin(), x.end());
  vector<pair<int, int>> a;
  for (auto i = (0); i < n; i++) {
    int pos = i;
    while (pos + 1 < n && x[pos + 1] == x[pos] + 1) pos++;
    a.push_back({x[i], x[pos]});
    i = pos;
  };
  vector<pair<int, pair<int, int>>> d;
  for (auto i = (1); i < (int)(a).size(); i++) {
    d.push_back({i, {a[i - 1].second + 1, a[i].first - 1}});
  };
  int cnt = 0;
  long long res = 0;
  vector<int> pos(m);
  int x0 = x[0], xn = x[n - 1];
  for (auto j = (1); j < m + 1; j++) {
    if (cnt >= m) break;
    if (cnt < m) {
      pos[cnt] = --x0;
      cnt++;
      res += j;
    }
    if (cnt < m) {
      pos[cnt] = ++xn;
      cnt++;
      res += j;
    }
    vector<int> f((int)(d).size());
    for (int i = 0; i < (int)(d).size();) {
      pair<int, int> in = d[i].second;
      int pp = d[i].first;
      if (cnt < m) {
        pos[cnt] = in.first;
        cnt++;
        in.first++;
        res += j;
      }
      if (in.first > in.second) {
        f[i++] = 1;
        continue;
      }
      if (cnt < m) {
        pos[cnt] = in.second;
        cnt++;
        in.second--;
        res += j;
      }
      if (in.first > in.second) {
        f[i++] = 1;
        continue;
      }
      if (cnt >= m) break;
      d[i] = (pair<int, pair<int, int>>){pp, (pair<int, int>)in};
      i++;
    }
    vector<pair<int, pair<int, int>>> dd;
    for (auto i = (0); i < (int)(d).size(); i++) {
      if (!f[i]) dd.push_back(d[i]);
    }
    d.clear();
    d.assign(dd.begin(), dd.end());
    ;
  }
  cout << res << endl;
  for (auto i = (0); i < m; i++) cout << pos[i] << " ";
  cout << endl;
}
