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 int maxn = 400000;
int n, m;
long long a[maxn];
pair<long long, long long> p[maxn];
long long solve(long long x) {
  long long ret = 0;
  for (int i = 0; i < 2 * n; i++) {
    ret += min(x, p[i].first);
  }
  return ret;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    if (!i) {
      p[i] = {m, -i - 1};
    } else {
      p[i] = {(a[i] - a[i - 1] - 1) / 2, -i - 1};
    }
    if (i == n - 1) {
      p[i + n] = {m, i + 1};
    } else {
      p[i + n] = {(a[i + 1] - a[i]) / 2, i + 1};
    }
  }
  long long l = 0, r = m;
  while (r - l > 1) {
    long long mid = (l + r) / 2;
    if (solve(mid) < m)
      l = mid;
    else
      r = mid;
  }
  long long ret = 0;
  vector<long long> ans;
  for (int i = 0, c = m - solve(l); i < 2 * n; i++) {
    int k = p[i].second < 0 ? -1 : 1;
    for (int j = 1; j <= min(l, p[i].first); j++) {
      ret += j;
      ans.push_back(a[k * p[i].second - 1] + k * j);
    }
    if (c && p[i].first >= r) {
      c--;
      ret += r;
      ans.push_back(a[k * p[i].second - 1] + k * r);
    }
  }
  sort(ans.begin(), ans.end());
  cout << ret << '\n';
  cout << ans[0];
  for (int i = 1; i < m; i++) cout << " " << ans[i];
  cout << '\n';
  return 0;
}
