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;
int inf = 1e9;
const int N = 2e5 + 10;
int n, m, cnt;
int a[N], ans[N];
map<int, int> mp;
set<int> st;
vector<int> v;
long long sum = 0;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    st.insert(a[i]);
    mp[a[i]] = 1;
  }
  sort(a + 1, a + 1 + n);
  int j = 1;
  for (j = 1;; j++) {
    bool f = 0;
    int y = 0;
    for (auto x : st) {
      if (mp[x - j] && mp[x + j]) {
        v.push_back(x);
        continue;
      }
      f = 0;
      if (!mp[x - j]) {
        cnt++;
        mp[x - j] = 1;
        ans[cnt] = x - j;
        sum += j;
      }
      if (!mp[x + j]) {
        cnt++;
        mp[x + j] = 1;
        ans[cnt] = x + j;
        sum += j;
      }
      y = x;
      if (cnt > m) {
        sum -= j;
        break;
      }
      if (cnt == m) break;
    }
    for (auto x : v) st.erase(x);
    v.clear();
    if (cnt >= m || st.size() == 0) break;
  }
  cout << sum << "\n";
  for (int i = 1; i <= m; i++) cout << ans[i] << " ";
}
