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 sum(long long a) {
  int sum = 0;
  while (a > 0) {
    sum = sum + (a % 10);
    a = a / 10;
  }
  return sum;
}
int count_digit(long long n) {
  int count = 0;
  while (n > 0) {
    n = n / 10;
    count++;
  }
  return count;
}
int binarySearch(int x, int y, long long z, long long v[]) {
  int low = x;
  int high = y;
  int mid = x + (y - x) / 2;
  while (low <= high) {
    if (v[mid] == z) return mid;
    if (v[mid] < z) return binarySearch(mid + 1, high, z, v);
    if (v[mid] > z) return binarySearch(low, mid - 1, z, v);
  }
  return -1;
}
long long modularExponentiation(long long x, long long n, long long M) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return modularExponentiation((x * x) % M, n / 2, M);
  else
    return (x * modularExponentiation((x * x) % M, (n - 1) / 2, M)) % M;
}
long long binaryExponentiation(long long x, long long n) {
  if (n == 0)
    return 1;
  else if (n % 2 == 0)
    return binaryExponentiation(x * x, n / 2);
  else
    return x * binaryExponentiation(x * x, (n - 1) / 2);
}
set<long long> s;
void genrate(long long n, int len, int max) {
  if (len > max) return;
  s.insert(n);
  genrate(n * 10 + 1, len + 1, max);
  genrate(n * 10 + 0, len + 1, max);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int tests = 1;
  while (tests--) {
    int n, m;
    cin >> n >> m;
    vector<long long> v(n);
    for (int i = 0; i < v.size(); i++) {
      cin >> v[i];
    }
    sort((v).begin(), (v).end());
    queue<pair<long long, long long>> Q;
    set<long long> s;
    for (int i = 0; i < v.size(); i++) {
      s.insert(v[i]);
      Q.push({v[i], 0});
    }
    long long res = 1;
    long long final = 0;
    vector<long long> ans;
    while (ans.size() < m) {
      int x = Q.size();
      for (int i = 0; i < x; i++) {
        if (ans.size() == m) break;
        pair<long long, long long> y = Q.front();
        Q.pop();
        if (y.second != 0) {
          ans.push_back(y.first);
          final = final + y.second;
        }
        if (s.find(y.first - 1) == s.end()) {
          s.insert(y.first - 1);
          Q.push({y.first - 1, y.second + 1});
        }
        if (s.find(y.first + 1) == s.end()) {
          s.insert(y.first + 1);
          Q.push({y.first + 1, y.second + 1});
        }
      }
    }
    cout << final << '\n';
    for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
  }
}
