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;
int n, m;
vector<int> drvece;
vector<int> l, r;
vector<int> nl, nr;
map<int, bool> bio;
vector<int> rj;
int main() {
  scanf("%d%d", &n, &m);
  int x;
  for (int i = 0; i < n; i++) {
    scanf("%d", &x);
    drvece.push_back(x);
    bio[x] = 1;
    l.push_back(x - 1);
    r.push_back(x + 1);
  }
  long long int dist = 0, br_ljudi = 0;
  long long int tr_dist = 1;
  bool z = 0;
  while (br_ljudi < m) {
    nl.clear();
    nr.clear();
    for (int i = 0; i < (int)l.size(); i++) {
      if (!bio[l[i]]) {
        dist += tr_dist;
        br_ljudi++;
        rj.push_back(l[i]);
        bio[l[i]] = 1;
        if (br_ljudi == m) {
          z = 1;
          break;
        }
        nl.push_back(l[i] - 1);
      }
    }
    if (z) break;
    for (int i = 0; i < (int)r.size(); i++) {
      if (!bio[r[i]]) {
        dist += tr_dist;
        br_ljudi++;
        rj.push_back(r[i]);
        bio[r[i]] = 1;
        if (br_ljudi == m) {
          z = 1;
          break;
        }
        nr.push_back(r[i] + 1);
      }
    }
    if (z) break;
    l = nl;
    r = nr;
    tr_dist++;
  }
  printf("%lld\n", dist);
  sort(rj.begin(), rj.end());
  for (int i = 0; i < (int)rj.size(); i++) {
    printf("%d ", rj[i]);
  }
  return 0;
}
