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 main() {
  long long int nTree, ppl;
  cin >> nTree >> ppl;
  vector<pair<long long int, long long int>> segments;
  vector<long long int> trees(nTree);
  for (long long int i = 0; i < nTree; i++) cin >> trees[i];
  sort(trees.begin(), trees.end());
  segments.push_back({trees[0], trees[0]});
  for (long long int i = 1; i < nTree; i++) {
    if (trees[i - 1] == trees[i] - 1)
      segments[segments.size() - 1].second = trees[i];
    else
      segments.push_back({trees[i], trees[i]});
  }
  long long int curDis = 1;
  long long int totalDis = 0;
  vector<long long int> yj(ppl);
  long long int curPpl = ppl;
  bool allConnected = false;
  for (long long int i = 0; i < ppl; i++) {
    segments[0].first -= 1;
    yj[ppl - curPpl] = segments[0].first;
    curPpl--;
    totalDis += curDis;
    if (curPpl == 0) break;
    bool tmp = false;
    for (long long int j = 0;
         j < segments.size() && curPpl > 0 && !allConnected; j++) {
      if (j > 0 && segments[j].first != segments[j - 1].second &&
          segments[j].first != segments[j - 1].second + 1) {
        segments[j].first -= 1;
        yj[ppl - curPpl] = segments[j].first;
        curPpl--;
        totalDis += curDis;
        tmp = true;
      }
      if (j < segments.size() - 1 &&
          segments[j].second != segments[j + 1].first && curPpl > 0 &&
          segments[j].second + 1 != segments[j + 1].first) {
        tmp = true;
        segments[j].second += 1;
        yj[ppl - curPpl] = segments[j].second;
        curPpl--;
        totalDis += curDis;
      }
    }
    if (!tmp) allConnected = true;
    if (curPpl == 0) break;
    segments[segments.size() - 1].second += 1;
    yj[ppl - curPpl] = segments[segments.size() - 1].second;
    curPpl--;
    totalDis += curDis;
    curDis++;
    if (curPpl == 0) break;
  }
  cout << totalDis << endl;
  for (long long int i = 0; i < yj.size(); i++) cout << yj[i] << " ";
  return 0;
}
