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>
#pragma GCC optimize(2)
using namespace std;
inline int read() {
  int x = 0, w = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    x = (x << 1) + (x << 3) + c - '0';
    c = getchar();
  }
  return w == 1 ? x : -x;
}
int n, m, pos[200005], cnt;
map<int, int> p;
vector<int> ans;
long long tmp;
set<int> s, tmp2;
int main() {
  n = read();
  m = read();
  for (int i = 1; i <= n; i++) pos[i] = read(), p[pos[i]] = 1, s.insert(pos[i]);
  for (int i = 1; cnt < m; i++) {
    tmp2 = s;
    for (set<int>::iterator it = s.begin(); it != s.end(); it++) {
      int tp = *it;
      if (p[tp + i] && p[tp - i])
        tmp2.erase(tp);
      else {
        if (p[tp + i] == 0)
          p[tp + i] = 1, cnt++, ans.push_back(tp + i), tmp += i;
        if (cnt == m) break;
        if (p[tp - i] == 0)
          p[tp - i] = 1, cnt++, ans.push_back(tp - i), tmp += i;
        if (cnt == m) break;
      }
    }
    s = tmp2;
  }
  cout << tmp << endl;
  for (int i = 0; i < ans.size(); i++) printf("%d ", ans[i]);
  return 0;
}
