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 ll = long long;
using ld = long double;
using namespace std;
const ll inf = 2e18;
const ll mod = 1e9 + 7;
const ll N = 2e5 + 5;
void code() {
  int n, m;
  cin >> n >> m;
  vector<int> a(n);
  for (auto &i : a) cin >> i;
  queue<pair<int, int>> q;
  set<int> occupied;
  for (auto i : a) q.emplace(i, 0), occupied.insert(i);
  long long dist = 0;
  int cnt = 0;
  vector<int> ans;
  while (!q.empty() and cnt < m) {
    auto [pos, d] = q.front();
    q.pop();
    int left = pos - 1;
    if (!occupied.count(left)) {
      q.emplace(left, d + 1);
      occupied.insert(left);
      cnt++;
      dist += d + 1;
      ans.push_back(left);
    }
    if (cnt >= m) break;
    int right = pos + 1;
    if (!occupied.count(right)) {
      q.emplace(right, d + 1);
      occupied.insert(right);
      cnt++;
      dist += d + 1;
      ans.push_back(right);
    }
  }
  cout << dist << "\n";
  sort((ans).begin(), (ans).end());
  for (auto i : ans) cout << i << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t = 1;
  while (t-- > 0) {
    code();
  }
}
