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;
inline void pisz(int n) { printf("%d\n", n); }
const int fx[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
const int fxx[8][2] = {{0, 1}, {0, -1}, {1, 0},  {-1, 0},
                       {1, 1}, {1, -1}, {-1, 1}, {-1, -1}};
template <typename T, typename TT>
ostream& operator<<(ostream& s, pair<T, TT> t) {
  return s << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, vector<T> t) {
  for (int i = 0; i < ((int)((t).size())); i++) s << t[i] << " ";
  return s;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  long long ans = 0;
  map<int, int> visited;
  map<int, int> distance;
  vector<int> x;
  vector<long long> y;
  queue<int> Q;
  for (int i = 0; i < n; i++) {
    int num;
    scanf("%d", &num);
    visited[num] = 1;
    distance[num] = 0;
    x.push_back(num);
    Q.push(num);
  }
  int i = 0;
  int te = 0;
  while (!Q.empty()) {
    te++;
    int cur = Q.front();
    Q.pop();
    if (distance[cur]) {
      ans += distance[cur];
      i++;
      y.push_back(cur);
    }
    if (i == m) break;
    if (!visited[cur - 1]) {
      distance[cur - 1] = distance[cur] + 1;
      Q.push(cur - 1);
      visited[cur - 1] = 1;
    }
    if (!visited[cur + 1]) {
      distance[cur + 1] = distance[cur] + 1;
      Q.push(cur + 1);
      visited[cur + 1] = 1;
    }
  }
  printf("%lld\n", ans);
  for (int i = 0; i <= y.size() - 1; i++)
    cout << y[i] << (i == y.size() - 1 ? '\n' : ' ');
}
