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 long long mod = 1000000007;
const long long INF = 100000010000;
long long n, m;
long long x[200010];
vector<long long> y;
long long ans;
long long gcd(long long x, long long y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long pow_mod(long long x, long long y) {
  if (y == 0) return 1;
  if (y % 2 == 1) {
    return x * pow_mod(x, y - 1) % mod;
  } else {
    long long half = pow_mod(x, y / 2);
    return half * half % mod;
  }
}
long long div_mod(long long x) { return pow_mod(x, mod - 2); }
void init() {
  cin >> n >> m;
  for (int i = 0; i < n; i++) {
    cin >> x[i];
  }
  sort(x, x + n);
}
void output() {
  cout << ans << endl;
  for (int i = 0; i < y.size(); i++) {
    cout << y[i];
    if (i == y.size() - 1) {
      cout << endl;
    } else {
      cout << ' ';
    }
  }
}
void solve() {
  vector<pair<long long, long long> > dif;
  for (int i = 1; i < n; i++) {
    if (x[i] - x[i - 1] == 1) continue;
    dif.push_back(pair<long long, long long>(x[i] - x[i - 1], i));
  }
  dif.push_back(pair<long long, long long>(INF, 0));
  dif.push_back(pair<long long, long long>(INF, n));
  sort(dif.begin(), dif.end());
  reverse(dif.begin(), dif.end());
  long long cnt = 0;
  for (int i = 1;; i++) {
    for (int j = 0; j < dif.size(); j++) {
      if (dif[j].first / 2 < i) {
        break;
      }
      if (dif[j].first % 2 == 0 && dif[j].first / 2 == i) {
        ans += i;
        if (dif[j].second == n) {
          y.push_back(x[n - 1] + i);
        } else {
          y.push_back(x[dif[j].second] - i);
        }
        if (y.size() == m) return;
      } else {
        if (dif[j].second == n) {
          ans += i;
          y.push_back(x[n - 1] + i);
          if (y.size() == m) return;
        } else if (dif[j].second == 0) {
          ans += i;
          y.push_back(x[0] - i);
          if (y.size() == m) return;
        } else {
          ans += i;
          y.push_back(x[dif[j].second - 1] + i);
          if (y.size() == m) return;
          ans += i;
          y.push_back(x[dif[j].second] - i);
          if (y.size() == m) return;
        }
      }
    }
  }
}
int main() {
  init();
  solve();
  output();
  long long N;
  cin >> N;
}
