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;
using ll = long long int;
const ll mod = 1000000007;
const ll M = 1000000000000000000ll;
inline ll add(ll a, ll b) { return ((a % mod + b % mod) % mod); }
inline ll sub(ll a, ll b) { return ((a % mod - b % mod + mod) % mod); }
inline ll mult(ll a, ll b) { return (((a % mod) * (b % mod)) % mod); }
inline ll power(ll n, ll x) {
  ll p = 1;
  while (x != 0) {
    if (x % 2 == 0) {
      n = mult(n, n);
      x = x / 2;
    }
    p = mult(p, n);
    x--;
  }
  return p;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  ll n, m, i;
  cin >> n >> m;
  vector<ll> v;
  set<ll> s;
  queue<pair<ll, ll>> q;
  for (i = 0; i < n; i++) {
    ll x;
    cin >> x;
    v.push_back(x);
    s.insert(x);
    q.push({x - 1, 1});
    q.push({x + 1, 1});
  }
  vector<ll> ans;
  ll mans = 0;
  ll c = 0;
  while (q.empty() == false) {
    ll t = q.front().first;
    if (s.find(t) == s.end()) {
      ll k = q.front().second;
      mans += k;
      ans.push_back(t);
      s.insert(t);
      q.push({t - 1, k + 1});
      q.push({t + 1, k + 1});
    }
    q.pop();
    if (ans.size() >= m) break;
  }
  cout << mans << "\n";
  for (i = 0; i < m; i++) cout << ans[i] << " ";
}
