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;
long long int n, m, i, j, k, l, arr[200009], brr[200009], crr[200009], a, b, c,
    d, e, f, g, h;
vector<pair<long long int, pair<long long int, long long int> > > v;
int main() {
  scanf("%lld %lld", &n, &m);
  k = 0;
  for (i = 0; i < n; i++) scanf("%lld", &arr[i]);
  sort(arr, arr + n);
  for (i = 1; i < n; i++) {
    l = arr[i] - arr[i - 1];
    v.push_back(make_pair(l, make_pair(arr[i], arr[i - 1])));
  }
  sort(v.begin(), v.end());
  a = 0;
  k = 0;
  l = 0;
  while (1) {
    b += 2;
    if (a + b >= m) break;
    for (j = k; j < n - 1 && a + b < m; j++) {
      d = v[j].first - (2 * c) - 1;
      if (d <= 0)
        k++;
      else if (d == 1) {
        brr[l++] = v[j].second.second + c + 1;
        k++;
        a++;
        h += c + 1;
      } else {
        brr[l++] = v[j].second.second + c + 1;
        a++;
        h += c + 1;
        if (a + b < m) {
          brr[l++] = v[j].second.first - c - 1;
          a++;
          h += c + 1;
        } else
          break;
      }
    }
    c++;
  }
  d = m - a;
  e = d / 2;
  f = d % 2;
  for (i = 0; i < e; i++) {
    brr[l++] = arr[0] - i - 1;
    h += i + 1;
  }
  for (i = 0; i < e; i++) {
    brr[l++] = arr[n - 1] + i + 1;
    h += i + 1;
  }
  if (f) {
    brr[l++] = arr[n - 1] + e + 1;
    h += e + 1;
  }
  sort(brr, brr + l);
  printf("%lld\n", h);
  for (i = 0; i < l; i++) printf("%lld ", brr[i]);
}
