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;
const int maxn = 2e5 + 10;
const long long inf = 1e15 + 7;
long long mod = 1e9 + 7;
int n, m;
map<int, int> vis;
vector<int> ans;
struct node {
  int x, tag;
} a[maxn];
queue<node> q;
long long bfs() {
  long long cnt = 0;
  while (!q.empty() && ans.size() < m) {
    node t = q.front(), next;
    q.pop();
    if (vis[t.x - 1] == 0) {
      cnt += 1LL * (t.tag + 1);
      next.x = t.x - 1;
      next.tag = t.tag + 1;
      vis[t.x - 1] = 1;
      q.push(next);
      ans.push_back(t.x - 1);
    }
    if (ans.size() == m) return cnt;
    if (vis[t.x + 1] == 0) {
      cnt += 1LL * (t.tag + 1);
      next.x = t.x + 1;
      next.tag = t.tag + 1;
      vis[t.x + 1] = 1;
      q.push(next);
      ans.push_back(t.x + 1);
    }
  }
  return cnt;
}
int main(void) {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i].x);
    vis[a[i].x] = 1;
    a[i].tag = 0;
    q.push(a[i]);
  }
  long long cnt = bfs();
  printf("%lld\n", cnt);
  for (int i = 0; i < m; i++) printf("%d%c", ans[i], i == m - 1 ? '\n' : ' ');
  return 0;
}
