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 int oo = 1e9 + 7;
const int modulo = 1e9 + 7;
template <typename T>
inline T sum(const T& a, const T& b) {
  return T((a % modulo + b % modulo) % modulo);
}
template <typename T>
inline T mul(const T& a, const T& b) {
  return T((1LL * (a % (1LL * modulo)) * (b % (1LL * modulo))) %
           (1LL * modulo));
}
string to_string(string a) { return "\"" + a + "\""; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  map<int, bool> vs;
  int n, m;
  queue<pair<int, int> > q;
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    int a;
    cin >> a;
    vs[a] = true;
    q.push(pair<int, int>(a, +1));
    q.push(pair<int, int>(a, -1));
  }
  long long ans = 0;
  vector<int> p;
  while (m) {
    pair<int, int> u = q.front();
    q.pop();
    if (vs.find(u.first + u.second) != vs.end()) continue;
    m--;
    ans += abs(u.second);
    p.push_back(u.first + u.second);
    vs[u.first + u.second] = true;
    for (int j = -1; j <= +1; ++j)
      if (j && vs.find(u.first + u.second + j) == vs.end()) {
        q.push(pair<int, int>(u.first, u.second + j));
      }
  }
  cout << ans << '\n';
  for (int v : p) cout << v << ' ';
  return 0;
}
