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;
long long infi = 2e9;
int main() {
  long long n, m;
  cin >> n >> m;
  set<vector<long long>> S;
  vector<long long> x(n), y;
  for (int i = 0; i < n; i++) cin >> x[i];
  sort(x.begin(), x.end());
  for (int i = 1; i < n; i++) {
    long long x1 = x[i - 1] + 1, x2 = (x[i - 1] + x[i]) / 2;
    if (x1 <= x2) {
      S.insert({1, 0, x1, x2});
    }
    long long x3 = ((x[i - 1] + x[i]) / 2) + 1, x4 = x[i] - 1;
    if (x4 >= x3) {
      S.insert({1, 1, x3, x4});
    }
  }
  S.insert({1, 1, -infi, x[0] - 1});
  S.insert({1, 0, x[n - 1] + 1, infi});
  long long ans = 0;
  for (int i = 0; i < m; i++) {
    auto it = S.begin();
    auto v = *it;
    long long a = v[0], b = v[1], c = v[2], d = v[3];
    ans += a;
    a += 1;
    S.erase(it);
    if (b == 0) {
      y.push_back(c);
      c += 1;
    } else {
      y.push_back(d);
      d -= 1;
    }
    if (c <= d) S.insert({a, b, c, d});
  }
  cout << ans << "\n";
  for (int i = 0; i < m; i++) cout << y[i] << " ";
}
