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;
long long int mi(long long int n, long long int m) {
  long long int pw = n;
  long long int ex = m - 2;
  long long int ans = 1;
  while (ex) {
    if (ex & 1) ans = ans * pw % m;
    pw = pw * pw % m;
    ex >>= 1;
  }
  return ans;
}
long long int pw(long long int a, long long int n) {
  long long int pw = a;
  long long int ex = n;
  long long int ans = 1;
  while (ex) {
    if (ex & 1) ans = ans * pw;
    pw = pw * pw;
    ex >>= 1;
  }
  return ans;
}
void show(long long int x) { cout << x << " "; }
long long int gcd(long long int a, long long int b) {
  return b == 0 ? a : gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return a / gcd(a, b) * b;
}
long long int max(long long int a, long long int b) { return a > b ? a : b; }
long long int min(long long int a, long long int b) { return a < b ? a : b; }
long long int Abs(long long int x) { return x > 0 ? x : (-x); }
bool comp(pair<int, int> a, pair<int, int> b) { return a.second < b.second; }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long int n, m;
  cin >> n >> m;
  vector<long long int> vec(n);
  for (int i = 0; i < n; i++) cin >> vec[i];
  sort(vec.begin(), vec.end());
  vector<pair<long long int, long long int> > dp(n);
  for (int i = 0; i < n; i++) {
    if (i == 0 || i == n - 1) {
      if (i == 0) {
        dp[i].first = vec[i] - 1;
        if (vec[i + 1] != vec[i] + 1)
          dp[i].second = vec[i] + 1;
        else
          dp[i].second = 1e18;
      } else {
        dp[i].second = vec[i] + 1;
        if (vec[i] - 1 != vec[i - 1])
          dp[i].first = vec[i] - 1;
        else
          dp[i].first = -(1e18);
      }
    } else {
      if (vec[i] + 1 != vec[i + 1])
        dp[i].second = vec[i] + 1;
      else
        dp[i].second = 1e18;
      if (vec[i] - 1 != vec[i - 1])
        dp[i].first = vec[i] - 1;
      else
        dp[i].first = -(1e18);
    }
  }
  map<long long int, bool> mp;
  for (long long int i : vec) mp[i] = true;
  set<int> st;
  for (int i = 0; i < n; i++) st.insert(i);
  vector<long long int> rk;
  long long int cnt = 0;
  long long int ans = 0;
  while (cnt < m) {
    vector<long long int> tmp;
    for (auto it = st.begin(); it != st.end(); it++) {
      int i = *it;
      if (dp[i].first != -(1e18)) {
        if (mp.count(dp[i].first))
          dp[i].first = -(1e18);
        else {
          mp[dp[i].first] = true;
          rk.push_back(dp[i].first);
          ans += Abs(vec[i] - dp[i].first);
          cnt++;
          if (mp.count(dp[i].first - 1))
            dp[i].first = -(1e18);
          else
            dp[i].first--;
          if (cnt == m) break;
        }
      }
      if (dp[i].second != 1e18) {
        if (mp.count(dp[i].second))
          dp[i].second = 1e18;
        else {
          mp[dp[i].second] = true;
          rk.push_back(dp[i].second);
          ans += Abs(dp[i].second - vec[i]);
          cnt++;
          if (mp.count(dp[i].second + 1))
            dp[i].second = 1e18;
          else
            dp[i].second++;
          if (cnt == m) break;
        }
      }
      if (dp[i].first == -(1e18) && dp[i].second == 1e18) tmp.push_back(i);
    }
    for (long long int i : tmp) st.erase(i);
  }
  cout << ans << "\n";
  for (long long int i : rk) show(i);
}
