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;
long long tr[200007];
int pe[200007];
long long sz[200007];
struct node {
  long long pos;
  long long sz;
};
stack<node> T;
queue<node> TT;
queue<int> q;
map<long long, long long> vis;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf("%lld", &tr[i]), vis[tr[i]] = 1;
  sort(tr + 1, tr + 1 + n);
  int j = 1;
  long long k;
  k = 0;
  while (j <= n) {
    if (j == 1 || j == n) TT.push({tr[j], 0});
    if (j == n) break;
    long long length = max(tr[j] - tr[j - 1] - 1, tr[j + 1] - tr[j] - 1);
    if (length > 0) TT.push({tr[j], length});
    j++;
  }
  int sum = 0;
  int len = 1;
  long long ans = 0;
  while (sum < m) {
    int times = TT.size();
    for (int i = 0; i < times; i++) {
      node now = TT.front();
      TT.pop();
      long long pos = now.pos;
      if (vis[pos + len] == 0) {
        vis[pos + len] = 1;
        if (sum < m) {
          ans += len;
          q.push(pos + len);
        }
        sum++;
      }
      if (vis[pos - len] == 0) {
        vis[pos - len] = 1;
        if (sum < m) {
          ans += len;
          q.push(pos - len);
        }
        sum++;
      }
      if (now.sz == 0 || len < now.sz) TT.push(now);
    }
    len++;
  }
  printf("%lld\n", ans);
  while (!q.empty()) {
    printf("%d ", q.front());
    q.pop();
  }
  return 0;
}
