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 + 7;
long long a[maxn];
long long b[maxn];
long long ans[maxn];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  a[0] = a[1] - m - m;
  a[n + 1] = a[n] + m + m;
  vector<pair<long long, long long> > v(n);
  for (int i = 0; i < n; i++) {
    v[i] = make_pair(a[i + 1], a[i + 1]);
  }
  int cnt = 0;
  long long dis = 1;
  long long now = 0;
  long long sum = 0;
  while (cnt < m) {
    int k = 0;
    now++;
    for (int i = 0; i < n; i++) {
      bool flag = 0;
      if (i - 1 < 0) {
        v[i].first -= dis;
        sum += now;
        if (dis > 0) ans[cnt++] = v[i].first;
        if (cnt >= m) break;
      } else {
        long long p = v[i].first - dis;
        if (p == v[i - 1].second) {
          flag = 1;
        } else {
          v[i].first -= dis;
          sum += now;
          if (dis > 0) ans[cnt++] = v[i].first;
          if (cnt >= m) break;
        }
      }
      if (i + 1 >= n) {
        v[i].second += dis;
        sum += now;
        if (dis > 0) ans[cnt++] = v[i].second;
        if (cnt >= m) break;
      } else {
        long long p = v[i].second + dis;
        if (p < v[i + 1].first) {
          v[i].second += dis;
          sum += now;
          if (dis > 0) ans[cnt++] = v[i].second;
          if (cnt >= m) break;
        }
      }
      if (flag == 0) {
        v[k] = v[i];
        k++;
      } else {
        v[k - 1].second = v[i].second;
      }
    }
    n = k;
    n = max(1, n);
  }
  cout << sum << endl;
  for (int i = 0; i < m; i++) cout << ans[i] << ' ';
  cout << endl;
  return 0;
}
