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;
struct S {
  int a, b;
  S() {}
  S(int _a, int _b) {
    a = _a;
    b = _b;
  }
  const bool operator<(const S &o) const { return a < o.a; }
};
string exm;
inline void exf(void) {
  cout << exm << "\n";
  exit(0);
}
template <typename T>
inline void showAll(vector<T> &v, string sep = "") {
  for (T &here : v) cout << here << sep;
}
template <typename T>
inline void showAll(T arr[], int st, int end, string sep = "") {
  for (int i = st; i <= end; i++) cout << arr[i] << sep;
}
template <typename T>
inline vector<int> int_seperation(T N, int d = 10) {
  vector<int> v;
  while (N) {
    v.push_back(N % d);
    N /= d;
  }
  reverse(v.begin(), v.end());
  return v;
}
const int SIZE = 200009;
int arr[SIZE];
set<int> used;
int main() {
  int n, k;
  scanf("%d %d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &arr[i]);
  }
  sort(arr + 1, arr + 1 + n);
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    q.push(arr[i]);
    used.insert(arr[i]);
  }
  while (k) {
    int cur = q.front();
    q.pop();
    int l = cur - 1, r = cur + 1;
    if (!used.count(l)) {
      used.insert(l);
      q.push(l);
      k--;
    }
    if (k == 0) break;
    if (!used.count(r)) {
      used.insert(r);
      q.push(r);
      k--;
    }
  }
  for (int i = 1; i <= n; i++) {
    used.erase(arr[i]);
  }
  long long sum = 0;
  for (int x : used) {
    auto it = lower_bound(arr + 1, arr + 1 + n, x);
    if (it == arr + 1 + n)
      sum += abs(x - arr[n]);
    else if (it == arr + 1)
      sum += abs(x - arr[1]);
    else {
      int idx = it - arr;
      sum += min(
          {abs(x - arr[idx]), abs(x - arr[idx + 1]), abs(x - arr[idx - 1])});
    }
  }
  printf("%lld\n", sum);
  for (int here : used) printf("%d ", here);
  return 0;
}
