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;
set<int> ocupados, arvores;
int vet[200001];
int main() {
  int N, K;
  scanf("%d %d", &N, &K);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &vet[i]);
    arvores.insert(vet[i]);
    ocupados.insert(vet[i]);
  }
  priority_queue<pair<int, int> > pq;
  for (int i = 1; i <= N; i++) {
    if (ocupados.count(vet[i] - 1) == 0) {
      pq.push({-1, vet[i] - 1});
      ocupados.insert(vet[i] - 1);
    }
    if (ocupados.count(vet[i] + 1) == 0) {
      pq.push({-1, vet[i] + 1});
      ocupados.insert(vet[i] + 1);
    }
  }
  int cont = 0;
  long long ans = 0;
  vector<int> pontos;
  while (cont != K) {
    auto p = pq.top();
    pq.pop();
    ans -= p.first;
    pontos.push_back(p.second);
    cont++;
    if (ocupados.count(p.second - 1) == 0) {
      auto it = arvores.lower_bound(p.second - 1);
      auto it2 = arvores.upper_bound(p.second - 1);
      long long d = 100000000000000LL;
      if (it != arvores.end()) d = min(d, abs(*it - (p.second - 1LL)));
      if (it2 != arvores.begin()) {
        it2--;
        d = min(d, abs(*it2 - (p.second - 1LL)));
      }
      pq.push({-d, p.second - 1});
      ocupados.insert(p.second - 1);
    }
    if (ocupados.count(p.second + 1) == 0) {
      auto it = arvores.lower_bound(p.second + 1LL);
      auto it2 = arvores.upper_bound(p.second + 1LL);
      long long d = 100000000000000LL;
      if (it != arvores.end()) d = min(d, abs(*it - (p.second + 1LL)));
      if (it2 != arvores.begin()) {
        it2--;
        d = min(d, abs(*it2 - (p.second + 1LL)));
      }
      pq.push({-d, p.second + 1});
      ocupados.insert(p.second + 1);
    }
  }
  printf("%lld\n", ans);
  for (auto p : pontos) printf("%d ", p);
  return 0;
}
