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 int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a * b / gcd(a, b);
}
long long int fexp(long long int a, long long int b) {
  long long int ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % 1000000007;
    b /= 2;
    a = a * a % 1000000007;
  }
  return ans;
}
long long int inverse(long long int a, long long int p) {
  return fexp(a, p - 2);
}
void ingraph(vector<vector<long long int> >& graph, long long int m) {
  long long int x, y;
  for (long long int i = 0; i < m; i++) {
    cin >> x >> y;
    x--, y--;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
}
long long int modify(long long int n) {
  long long int res = n;
  res %= 1000000007;
  res += 1000000007;
  res %= 1000000007;
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  t = 1;
  while (t--) {
    long long int n, m;
    cin >> n >> m;
    set<long long int> xp, all;
    map<long long int, long long int> r, l;
    long long int x;
    for (long long int i = 0; i < n; i++) {
      cin >> x;
      xp.insert(x), all.insert(x);
      r[x] = 1;
      l[x] = 1;
    }
    long long int len = 1;
    long long int res = 0;
    set<long long int> y;
    while (y.size() < m) {
      set<long long int> tmp;
      tmp = xp;
      for (auto& v : tmp) {
        if (y.size() >= m) break;
        if (l[v] && all.find(v - len) == all.end() &&
            y.find(v - len) == y.end()) {
          y.insert(v - len);
          res += len;
        } else
          l[v] = 0;
        if (y.size() >= m) break;
        if (r[v] && all.find(v + len) == all.end() &&
            y.find(v + len) == y.end()) {
          y.insert(v + len);
          res += len;
        } else
          r[v] = 0;
        if (l[v] == 0 && r[v] == 0) xp.erase(v);
      }
      len++;
      if (y.size() >= m) break;
    }
    set<long long int>::iterator itr = y.begin();
    cout << res << "\n";
    while (itr != y.end()) {
      cout << *itr << " ";
      itr++;
    }
  }
}
