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;
template <typename T>
int sign(T a) {
  if (a == 0) return 0;
  return (a > 0 ? 1 : -1);
}
template <typename T1, typename T2>
T2 mod(T1 x, T2 MOD) {
  if (x < 0) x += MOD;
  if (x >= MOD) x -= MOD;
  return x;
}
template <typename T>
T sqr(T a) {
  return a * a;
}
template <typename T>
bool uax(T& a, const T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T>
bool uin(T& a, const T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& p) {
  in >> p.first >> p.second;
  return in;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& out, pair<T1, T2>& p) {
  out << '(' << p.first << ' ' << p.second << ')';
  return out;
}
mt19937_64 gen_rand;
const int N = (int)2e5 + 7;
const int LOG = 31;
const int N_ = (int)5007;
const int MAXN = (int)8e5 + 7;
const int MOD = (int)998244353;
const int INF = (int)1e8 + 7;
const int CONST = 5000;
const long long LINF = (int64_t)1e18 + 10;
const long double PI = 3.1415926535897932384626433832795;
const long double EPS = 5e-1;
int n, m, x[N];
map<int, int> dist;
int main() {
  cout << fixed << setprecision(20);
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  gen_rand.seed(time(0));
  cin >> n >> m;
  queue<int> q;
  for (int i = 0; i < n; i++) {
    cin >> x[i];
    dist[x[i]] = 0;
    q.push(x[i]);
  }
  long long sum = 0;
  vector<int> ans;
  while ((int)ans.size() != m) {
    int x = q.front();
    q.pop();
    if (dist[x] > 0) {
      ans.push_back(x);
      sum += dist[x];
    }
    if (!dist.count(x - 1)) {
      dist[x - 1] = dist[x] + 1;
      q.push(x - 1);
    }
    if (!dist.count(x + 1)) {
      dist[x + 1] = dist[x] + 1;
      q.push(x + 1);
    }
  }
  cout << sum << endl;
  for (int x : ans) cout << x << ' ';
}
