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 double pi = 3.1415926535;
void FAST() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
bool isprime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
vector<long long> digit(long long n) {
  long long arr[10000];
  long long i = 0;
  long long j, r;
  vector<long long> v;
  while (n != 0) {
    r = n % 10;
    arr[i] = r;
    i++;
    n = n / 10;
  }
  for (j = i - 1; j > -1; j--) {
    v.push_back(arr[j]);
  }
  return v;
}
long long sum(long long x) {
  long long ans = 0;
  while (x > 0) {
    ans += (x % 10);
    x /= 10;
  }
  return ans;
}
bool opp(int x, int y) { return ((x ^ y) < 0); }
bool bysec(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second < b.second);
}
int main() {
  FAST();
  long long n, k;
  cin >> n >> k;
  long long arr[n];
  map<long long, long long> m;
  queue<long long> q;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    m[arr[i]] = 0;
    q.push(arr[i]);
  }
  vector<long long> answ;
  long long ans = 0;
  while (!q.empty()) {
    if (int(answ.size()) == k) break;
    int curr = q.front();
    q.pop();
    if (m[curr] != 0) {
      ans += m[curr];
      answ.push_back(curr);
    }
    if (m.count(curr - 1) != 1) {
      m[curr - 1] = m[curr] + 1;
      q.push(curr - 1);
    }
    if (m.count(curr + 1) != 1) {
      m[curr + 1] = m[curr] + 1;
      q.push(curr + 1);
    }
  }
  cout << ans << '\n';
  for (auto i : answ) cout << i << " ";
  cout << '\n';
}
