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 mod = 1e9 + 7;
const int MAXN = 100010;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long i, j, k, u, n, m, t, x, z, v, cnt, a[1000005], y, tot = 0, maxx, ans,
                                                            minn, flag;
string len;
queue<long long> que;
vector<long long> vx;
map<long long, long long> mp;
int main() {
  cin >> n >> m;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    mp[a[i]] = 1;
  }
  ans = 0;
  for (i = 1; i <= n; i++)
    if (!mp[a[i] - 1] || !mp[a[i] + 1]) que.push(a[i]);
  while (!que.empty()) {
    long long now = que.front();
    que.pop();
    if (!mp[now - 1]) {
      mp[now - 1] = mp[now] + 1;
      ans += mp[now];
      if (!mp[now - 2]) que.push(now - 1);
      vx.push_back(now - 1);
      m--;
    }
    if (!m) break;
    if (!mp[now + 1]) {
      mp[now + 1] = mp[now] + 1;
      ans += mp[now];
      if (!mp[now + 2]) que.push(now + 1);
      vx.push_back(now + 1);
      m--;
    }
    if (!m) break;
  }
  cout << ans << endl;
  for (auto i : vx) cout << i << " ";
  return 0;
}
