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;
void debug(int* a, int n, bool flag = false) {
  if (!flag) printf("array ");
  for (int i = 1; i <= n; i++) printf("%d ", a[i]);
  printf("\n");
}
void debug(vector<int>& E, bool flag = false) {
  if (!flag) printf("vector ");
  for (auto it : E) printf("%d ", it);
  printf("\n");
}
const int maxn = 2e5 + 5;
int a[maxn];
set<pair<int, int> > s;
int n, m;
map<int, bool> Map;
vector<int> res;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    Map[a[i]] = true;
    s.insert(make_pair(i, -1));
    s.insert(make_pair(i, 1));
  }
  sort(a + 1, a + 1 + n);
  long long ans = 0;
  long long k = 1;
  while (m) {
    auto it = s.begin();
    while (it != s.end()) {
      long long val = a[it->first] + k * it->second;
      if (!Map.count(val)) {
        res.push_back(val);
        ans += k;
        it++;
        m--;
        Map[val] = true;
        if (!m) break;
      } else
        s.erase(it++);
    }
    k++;
  }
  printf("%I64d\n", ans);
  debug(res, true);
  return 0;
}
