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>
inline long long read() {
  long long f = 1;
  long long x = 0;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    x = (x << 3) + (x << 1) + ch - '0', ch = getchar();
  return x * f;
}
using namespace std;
void print(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) print(x / 10);
  putchar(x % 10 + '0');
}
struct node {
  int x, from;
};
queue<node> q;
map<int, bool> mp;
vector<int> vec;
int n, m;
int dir[2]{-1, 1};
long long bfs() {
  long long ans = 0;
  while (!q.empty()) {
    node st = q.front();
    q.pop();
    for (int i = 0; i < 2; i++) {
      node ne = {st.x + dir[i], st.from};
      if (mp[ne.x] == 0) {
        q.push(ne);
        vec.push_back(ne.x);
        mp[ne.x] = 1;
        ans += abs(ne.x - ne.from);
        if (vec.size() == m) return ans;
      }
    }
  }
}
int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) {
    int x;
    x = read();
    mp[x] = 1;
    q.push({x, x});
  }
  print(bfs());
  puts("");
  int size = vec.size();
  for (int i = 0; i < size; i++) {
    print(vec[i]);
    putchar(' ');
  }
}
