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;
int n, m, q;
int a[200009];
set<pair<int, int> > s;
set<int> se;
bool check(int len) {
  int ret = 0;
  for (int i = 0; i < n; i++) {
    auto it = se.find(a[i]);
    if (it != se.begin()) {
      it--;
      ret += min(len, (a[i] - (*it) - 1) / 2);
      it++;
    } else
      ret += len;
    it++;
    if (it != se.end()) {
      ret += min(len, ((*it) - a[i] - 1) / 2 + ((*it) - a[i] - 1) % 2);
    } else
      ret += len;
  }
  return (ret >= m);
}
long long sol(int len) {
  long long ret = 0;
  for (int i = 0; i < n; i++) {
    auto it = se.find(a[i]);
    int st = a[i] - len;
    if (it != se.begin()) {
      it--;
      st = max(st, a[i] - (a[i] - (*it) - 1) / 2);
      it++;
    }
    for (int j = st; j < a[i]; j++) {
      s.insert({a[i] - j, j});
      ret += a[i] - j;
    }
    int en = a[i] + len;
    it++;
    if (it != se.end()) {
      en = min(en, a[i] + (((*it) - a[i] - 1) / 2 + ((*it) - a[i] - 1) % 2));
    }
    for (int j = a[i] + 1; j <= en; j++) {
      s.insert({j - a[i], j});
      ret += j - a[i];
    }
  }
  int sz = s.size() - m;
  auto it = s.end();
  while (sz--) it--, ret -= it->first;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0), cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    se.insert(a[i]);
  }
  int l = 0, r = 2e5;
  while (r - l > 1) {
    if (check((l + r) / 2))
      r = (l + r) / 2;
    else
      l = (l + r) / 2;
  }
  cout << sol(r) << endl;
  for (auto it = s.begin(); it != s.end(); it++) {
    cout << it->second << " ", q++;
    if (q == m) break;
  }
  cout << endl;
}
