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;
const long long INF = 1e9 + 7;
const long long sz = 1e5 + 20;
const long long N = 1e5 + 20;
const long long big = 2 * 1e9;
const long long small = -big;
bool safe(long long x) {
  if (x >= small and x <= big) return true;
  return false;
}
void solve() {
  long long n, m;
  cin >> n >> m;
  map<long long, long long> vis;
  long long tree[n];
  for (long long i = 0; i < n; i++) cin >> tree[i], vis[tree[i]] = 1;
  queue<pair<long long, long long> > q;
  for (long long i = 0; i < n; i++) q.push(make_pair(tree[i], 0));
  long long cnt = 0;
  vector<long long> v;
  long long s = 0;
  while (!q.empty() and cnt <= m) {
    pair<long long, long long> fro = q.front();
    q.pop();
    long long dist = fro.second + 1;
    long long fr = fro.first;
    if (!vis[fr - 1] and safe(fr - 1)) {
      q.push(make_pair(fr - 1, dist));
      cnt++;
      vis[fr - 1] = true;
      v.push_back(fr - 1);
      s += dist;
    }
    if (cnt == m) break;
    if (!vis[fr + 1] and safe(fr + 1)) {
      q.push(make_pair(fr + 1, dist));
      cnt++;
      vis[fr + 1] = true;
      v.push_back(fr + 1);
      s += dist;
    }
    if (cnt == m) break;
  }
  cout << s << endl;
  for (auto x : v) cout << x << " ";
  cout << endl;
  return;
}
int32_t main() {
  long long t = 1;
  while (t--) solve();
}
