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 mxn = 505;
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<long long> a(n);
  for (long long& x : a) cin >> x;
  multiset<pair<long long, pair<long long, long long>>> f;
  reverse(a.begin(), a.end());
  a.push_back(-1e11);
  reverse(a.begin(), a.end());
  a.push_back(1e11);
  sort(a.begin(), a.end());
  for (long long i = 1; i <= n; i++) {
    pair<long long, pair<long long, long long>> ff =
        make_pair(1LL, make_pair((a[i] - a[i - 1]) / 2, i * -1));
    pair<long long, pair<long long, long long>> ss =
        make_pair(1LL, make_pair((a[i + 1] - a[i] - 1) / 2, i * 1));
    if (ff.second.first) f.insert(ff);
    if (ss.second.first) f.insert(ss);
  }
  long long ans = 0;
  vector<long long> vans;
  while (m--) {
    auto it = f.begin();
    pair<long long, pair<long long, long long>> ch = *it;
    f.erase(ch);
    ans += ch.first;
    vans.push_back(a[abs(ch.second.second)] +
                   ch.first * (ch.second.second / abs(ch.second.second)));
    --ch.second.first, ++ch.first;
    if (ch.second.first == 0) continue;
    f.insert(ch);
  }
  cout << ans;
  sort(vans.begin(), vans.end());
  cout << endl;
  for (long long x : vans) cout << x << " ";
}
signed main() {
  long long t = 1;
  while (t--) {
    solve();
    cout << endl;
  }
}
