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 long long INF = 2000000000000000000LL;
const int inf = 0x3f3f3f3f;
const long double EPS = 1e-9;
struct node {
  int root;
  int move;
  int cost = 0;
};
int arr[200009];
void bfs(int n, int m) {
  map<int, bool> origine, taken;
  std::vector<int> res;
  queue<node> Q;
  node nxtmove, now;
  int left, right, cst = 0;
  for (int i = 0; i < n; i++) origine[arr[i]] = 1;
  for (int i = 0; i < n; i++) {
    cst = 1, left = arr[i] + cst, right = arr[i] - cst;
    if (taken.find(left) == taken.end() &&
        origine.find(left) == origine.end()) {
      nxtmove.root = arr[i], nxtmove.move = left, nxtmove.cost = 1;
      Q.push(nxtmove);
      taken[left] = 1;
    }
    if (taken.find(right) == taken.end() &&
        origine.find(right) == origine.end()) {
      nxtmove.root = arr[i], nxtmove.move = right, nxtmove.cost = 1;
      Q.push(nxtmove);
      taken[right] = 1;
    }
  }
  long long cost = 0;
  while (!Q.empty() && m) {
    now = Q.front();
    Q.pop();
    if (m) m--, cost += now.cost, res.push_back(now.move);
    left = now.move - 1, right = now.move + 1, now.cost++;
    if (taken.find(left) == taken.end() &&
        origine.find(left) == origine.end()) {
      nxtmove.root = now.root, nxtmove.move = left, nxtmove.cost = now.cost;
      Q.push(nxtmove);
      taken[left] = 1;
    }
    if (taken.find(right) == taken.end() &&
        origine.find(right) == origine.end()) {
      nxtmove.root = now.root, nxtmove.move = right, nxtmove.cost = now.cost;
      Q.push(nxtmove);
      taken[right] = 1;
    }
  }
  cout << cost << endl;
  for (auto it : res) cout << it << " ";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  bfs(n, m);
  cout << endl;
  return 0;
}
