#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
map<long long, long long> mp;
struct point {
  int pos, val;
  bool operator<(const point &x) const { return val > x.val; }
};
long long ans[200005], a[200005];
priority_queue<point> q;
int main() {
  long long n, m, t, sum, k;
  point temp;
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    mp[a[i]]++;
  }
  for (long long i = 1; i <= n; i++) {
    if (mp[a[i] - 1] == 0) {
      temp.pos = a[i] - 1;
      temp.val = 1;
      q.push(temp);
      mp[temp.pos]++;
    }
    if (mp[a[i] + 1] == 0) {
      temp.pos = a[i] + 1;
      temp.val = 1;
      q.push(temp);
      mp[temp.pos]++;
    }
  }
  sum = 0;
  k = 0;
  while (m--) {
    temp = q.top();
    q.pop();
    ans[++k] = temp.pos;
    sum += temp.val;
    point Next;
    if (mp[temp.pos - 1] == 0) {
      Next.pos = temp.pos - 1;
      Next.val = temp.val + 1;
      q.push(Next);
      mp[Next.pos]++;
    }
    if (mp[temp.pos + 1] == 0) {
      Next.pos = temp.pos + 1;
      Next.val = temp.val + 1;
      q.push(Next);
      mp[Next.pos]++;
    }
  }
  printf("%lld\n", sum);
  for (long long i = 1; i <= k; i++) {
    printf("%lld", ans[i]);
    if (i == k)
      printf("\n");
    else
      printf(" ");
  }
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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 

### Response

```cpp
#include <bits/stdc++.h>
#pragma warning(disable : 4996)
using namespace std;
map<long long, long long> mp;
struct point {
  int pos, val;
  bool operator<(const point &x) const { return val > x.val; }
};
long long ans[200005], a[200005];
priority_queue<point> q;
int main() {
  long long n, m, t, sum, k;
  point temp;
  scanf("%lld%lld", &n, &m);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    mp[a[i]]++;
  }
  for (long long i = 1; i <= n; i++) {
    if (mp[a[i] - 1] == 0) {
      temp.pos = a[i] - 1;
      temp.val = 1;
      q.push(temp);
      mp[temp.pos]++;
    }
    if (mp[a[i] + 1] == 0) {
      temp.pos = a[i] + 1;
      temp.val = 1;
      q.push(temp);
      mp[temp.pos]++;
    }
  }
  sum = 0;
  k = 0;
  while (m--) {
    temp = q.top();
    q.pop();
    ans[++k] = temp.pos;
    sum += temp.val;
    point Next;
    if (mp[temp.pos - 1] == 0) {
      Next.pos = temp.pos - 1;
      Next.val = temp.val + 1;
      q.push(Next);
      mp[Next.pos]++;
    }
    if (mp[temp.pos + 1] == 0) {
      Next.pos = temp.pos + 1;
      Next.val = temp.val + 1;
      q.push(Next);
      mp[Next.pos]++;
    }
  }
  printf("%lld\n", sum);
  for (long long i = 1; i <= k; i++) {
    printf("%lld", ans[i]);
    if (i == k)
      printf("\n");
    else
      printf(" ");
  }
  return 0;
}

```