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;
using LL = int64_t;
const int maxn = 2e5 + 5;
vector<pair<LL, LL>> mp;
vector<LL> res;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    LL x;
    cin >> x;
    mp.push_back({x, x});
  }
  sort(mp.begin(), mp.end());
  LL ans = 0, num = 1;
  int sz = 0;
  while (sz < m) {
    for (int i = 0; i < n; i++) {
      LL x = mp[i].first - 1;
      if (i == 0 || mp[i].first > mp[i - 1].second + 1) {
        res.push_back(x);
        ans += num;
        sz++;
        mp[i].first--;
      }
      if (sz == m) break;
      x = mp[i].second + 1;
      if (i == n - 1 || mp[i].second + 1 < mp[i + 1].first) {
        res.push_back(x);
        ans += num;
        sz++;
        mp[i].second++;
      }
      if (sz == m) break;
    }
    if (sz == m) break;
    num++;
    pair<LL, LL> now = mp[0];
    int id = 0;
    for (int i = 1; i < n; i++) {
      if (mp[i].first == now.second + 1)
        now.second = mp[i].second;
      else {
        mp[id] = now;
        id++;
        now = mp[i];
      }
    }
    if (mp[id - 1] != now) {
      mp[id] = now;
      id++;
    }
    n = id;
  }
  cout << ans << '\n';
  for (auto it : res) cout << it << ' ';
  return 0;
}
