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 mx(vector<int> a) {
  vector<int> cnt(a.size() + 3, 0);
  for (auto i : a) cnt[i]++;
  a = cnt;
  for (int i = 1; i < a.size() - 1; i++) {
    if (a[i] == 0) continue;
    if (a[i - 1] == 0) {
      a[i - 1]++;
      a[i]--;
    }
    if (a[i] > 1) {
      a[i]--;
      a[i + 1]++;
    }
  }
  int ans = 0;
  for (auto i : a) ans += (i != 0);
  return ans;
}
int mn(vector<int> a) {
  sort((a).begin(), (a).end());
  a.erase(unique((a).begin(), (a).end()), a.end());
  int n = (int)(a).size();
  vector<vector<int>> dp(3, vector<int>(n, n));
  dp[0][0] = dp[1][0] = dp[2][0] = 1;
  for (int i = 1; i < n; i++) {
    for (int p = 0; p < 3; p++) {
      for (int c = 0; c < 3; c++) {
        int cur = a[i] + 1 - p;
        int prev = a[i - 1] + 1 - c;
        dp[p][i] = min(dp[p][i], dp[c][i - 1] + (cur != prev));
      }
    }
  }
  return min({dp[0][n - 1], dp[1][n - 1], dp[2][n - 1]});
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  long long s = 0;
  vector<int> poses;
  map<int, int> used;
  queue<pair<int, int>> q;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    used[x] = 1;
    q.push({1, x + 1});
    q.push({-1, x - 1});
  }
  while ((int)(poses).size() < m) {
    auto cur = q.front();
    q.pop();
    if (used[cur.second]) {
      continue;
    }
    used[cur.second] = 1;
    poses.push_back(cur.second);
    s += abs(cur.first);
    if (cur.first < 0) {
      q.push({cur.first - 1, cur.second - 1});
    } else {
      q.push({cur.first + 1, cur.second + 1});
    }
  }
  cout << s << '\n';
  for (int i : poses) cout << i << ' ';
  return 0;
}
