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>
#pragma GCC optimize("O3")
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const long long MOD = 1e9 + 7;
const long long N = 2e5 + 10;
const long long INF = 1e9 + 10;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long n, m;
  cin >> n >> m;
  vector<long long> v(n);
  set<int> s;
  set<int> tree;
  map<int, int> dist;
  queue<long long> q;
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    dist[v[i]] = 0;
    s.insert(v[i]);
    tree.insert(v[i]);
    q.push(v[i]);
  }
  long long ans = 0;
  while (not q.empty()) {
    if (m <= 0) break;
    long long f = q.front();
    q.pop();
    if (s.find(f - 1) == s.end() && m > 0) {
      dist[f - 1] = dist[f] + 1;
      ans += dist[f - 1];
      s.insert(f - 1);
      q.push(f - 1);
      m--;
    }
    if (s.find(f + 1) == s.end() && m > 0) {
      dist[f + 1] = dist[f] + 1;
      ans += dist[f + 1];
      s.insert(f + 1);
      q.push(f + 1);
      m--;
    }
  }
  cout << ans << endl;
  for (long long c : s) {
    if (tree.find(c) == tree.end()) {
      cout << c << ' ';
    }
  }
  return 0;
}
