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 max(long long a, long long b) {
  if (a > b) {
    return a;
  } else {
    return b;
  }
}
struct node {
  long long x;
  long long l;
  long long r;
};
bool comp(node a, node b) { return a.x < b.x; }
bool cmp(node a, node b) {
  return max(a.r - a.x, a.x - a.l) < max(b.r - b.x, b.x - b.l);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    long long n, m;
    cin >> n >> m;
    node v[n];
    for (long long i = 0; i < n; i++) {
      cin >> v[i].x;
    }
    sort(v, v + n, comp);
    v[0].l = -2000000000;
    for (long long i = 1; i < n; i++) {
      v[i].l = v[i - 1].x + (v[i].x - v[i - 1].x + 1) / 2;
    }
    for (long long i = 0; i < n - 1; i++) {
      v[i].r = v[i + 1].l - 1;
    }
    v[n - 1].r = 2000000000;
    sort(v, v + n, cmp);
    long long ll[n];
    long long rl[n];
    for (long long i = 0; i < n; i++) {
      ll[i] = v[i].x - 1;
      rl[i] = v[i].x + 1;
    }
    long long b[m];
    long long ctr = 0;
    long long ans = 0;
    long long val = 0;
    long long pos = 0;
    while (ctr != m) {
      val++;
      for (long long i = pos; i < n; i++) {
        if (ll[i] < v[i].l && rl[i] > v[i].r) {
          pos++;
          continue;
        }
        if (ll[i] >= v[i].l) {
          ans = ans + val;
          b[ctr] = ll[i];
          ll[i]--;
          ctr++;
        }
        if (ctr == m) {
          break;
        }
        if (rl[i] <= v[i].r) {
          ans = ans + val;
          b[ctr] = rl[i];
          rl[i]++;
          ctr++;
        }
        if (ctr == m) {
          break;
        }
      }
      if (ctr == m) {
        break;
      }
    }
    cout << ans << endl;
    for (long long i = 0; i < m; i++) {
      cout << b[i] << " ";
    }
  }
}
