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;
bool isv(char x) {
  if (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u') {
    return true;
  }
  return false;
}
bool isp(long long n) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
string tobin(long long n) {
  string res = "";
  long long i = 0;
  while (n > 0) {
    if (n % 2 == 1) {
      res = res + '1';
    } else {
      res = res + '0';
    }
    n /= 2;
    i++;
  }
  reverse(res.begin(), res.end());
  return res;
}
long long todec(string s) {
  long long n = s.size(), ans = 0, temp = 1;
  for (long long i = n - 1; i >= 0; i--) {
    long long x = s[i] - '0';
    ans += x * temp;
    temp *= 2;
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio();
  cin.tie(0);
  cout.tie(0);
  {
    long long p, t;
    cin >> t >> p;
    long long a[t];
    map<long long, long long> vis;
    list<long long> q;
    for (long long i = 0; i < t; i++) {
      cin >> a[i];
      vis[a[i]] = 0;
      q.push_back(a[i]);
    }
    vector<long long> ans;
    long long dist = 0;
    while (!q.empty()) {
      if (ans.size() == p) break;
      long long temp = q.front();
      q.pop_front();
      if (vis[temp] != 0) {
        dist += vis[temp];
        ans.push_back(temp);
      }
      if (!vis.count(temp - 1)) {
        vis[temp - 1] = vis[temp] + 1;
        q.push_back(temp - 1);
      }
      if (!vis.count(temp + 1)) {
        vis[temp + 1] = vis[temp] + 1;
        q.push_back(temp + 1);
      }
    }
    cout << dist << endl;
    for (long long i = 0; i < p; i++) {
      cout << ans[i] << " ";
    }
    cout << endl;
    ;
  }
  return 0;
}
