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 N = 200005;
struct node {
  long long first, second;
};
long long n, k;
long long a[N], ans[N];
map<long long, long long> vis;
queue<node> q;
signed main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  long long kk = k, res = 0;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    vis[a[i]] = 1;
  }
  for (long long i = 1; i <= n; i++) {
    if (!vis[a[i] - 1]) q.push({a[i] - 1, 1}), vis[a[i] - 1] = 1;
    if (!vis[a[i] + 1]) q.push({a[i] + 1, 1}), vis[a[i] + 1] = 1;
  }
  while (kk) {
    node now = q.front();
    q.pop();
    if (!vis[now.first - 1])
      q.push({now.first - 1, now.second + 1}), vis[now.first - 1] = 1;
    if (!vis[now.first + 1])
      q.push({now.first + 1, now.second + 1}), vis[now.first + 1] = 1;
    ans[kk] = now.first;
    res += now.second;
    kk--;
  }
  cout << res << endl;
  for (long long i = 1; i <= k; i++) cout << ans[i] << " ";
}
