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;
long long power(long long, long long);
long long mpower(long long, long long);
long long BIT[1000009];
long long query(long long);
void update(long long, long long);
long long n, m, t;
long long ans[1000009];
set<long long> s;
map<long long, long long> dis;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  queue<long long> q;
  for (long long i = 1; i <= n; i++) {
    cin >> t;
    q.push(t);
    s.insert(t);
  }
  long long cnt = 0, a = 0;
  while (!q.empty() and cnt < m) {
    long long u = q.front();
    q.pop();
    if (s.find(u + 1) == s.end()) {
      ans[cnt++] = u + 1;
      q.push(u + 1);
      dis[u + 1] = dis[u] + 1;
      if (cnt <= m) a += dis[u + 1];
      s.insert(u + 1);
    }
    if (s.find(u - 1) == s.end()) {
      ans[cnt++] = u - 1;
      q.push(u - 1);
      dis[u - 1] = dis[u] + 1;
      if (cnt <= m) a += dis[u - 1];
      s.insert(u - 1);
    }
  }
  cout << a << endl;
  for (long long i = 0; i < m; i++) cout << ans[i] << " ";
  return 0;
}
void update(long long x, long long delta) {
  for (; x < 1000009; x += x & -x) BIT[x] += delta;
}
long long query(long long x) {
  long long sum = 0;
  for (; x > 0; x -= x & -x) sum += BIT[x];
  return sum;
}
long long power(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 == 1) res = res * a;
    a = a * a;
    b = b / 2;
  }
  return res;
}
long long mpower(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b % 2 == 1) {
      res = (res * a) % 1000000007;
    }
    a = (a * a) % 1000000007;
    b = b / 2;
  }
  return res;
}
