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;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
long long exp(long long x, long long y) {
  x %= 1000000007;
  long long res = 1;
  while (y) {
    if (y & 1) res = res * x % 1000000007;
    x = x * x % 1000000007;
    y >>= 1;
  }
  return res;
}
long long fact(long long n) {
  long long res = 1;
  for (long long i = 2; i <= n; ++i) (res *= i) %= 1000000007;
  return res;
}
bool isprime(long long n) {
  if (n <= 1) return 0;
  for (long long i = 2; i * i <= n; ++i)
    if (n % i == 0) return 0;
  return 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<long long> v(n);
  for (int i = 0; i < n; ++i) cin >> v[i];
  sort(v.begin(), v.end());
  long long l = 0, r = m;
  while (r > l + 1) {
    long long mid = (l + r) / 2, cnt = 2 * mid;
    for (int i = 1; i < n; ++i) {
      cnt += min(2 * mid, v[i] - v[i - 1] - 1);
    }
    if (cnt >= m)
      r = mid;
    else
      l = mid;
  }
  vector<int> ans;
  long long p = 0;
  for (int i = 0; i < n; ++i) {
    if (i && v[i] - v[i - 1] - 1 <= 2 * (r - 1)) {
      for (int j = v[i - 1] + 1; j <= v[i] - 1; ++j) {
        ans.push_back(j);
        p += min(j - v[i - 1], v[i] - j);
      }
    } else {
      if (i)
        for (int j = v[i - 1] + 1; j <= v[i - 1] + r - 1; ++j) {
          ans.push_back(j);
          p += j - v[i - 1];
        }
      for (int j = v[i] - r + 1; j <= v[i] - 1; ++j) {
        ans.push_back(j);
        p += v[i] - j;
      }
    }
  }
  for (int j = v[n - 1] + 1; j <= v[n - 1] + r - 1; ++j) {
    ans.push_back(j);
    p += j - v[n - 1];
  }
  m -= int(ans.size());
  p += m * r;
  for (int i = 0; i < n; ++i) {
    if (i == 0) {
      if (m > 0) {
        ans.push_back(v[i] - r);
        m -= 1;
      }
    } else {
      int rem = max(0ll, v[i] - v[i - 1] - 1 - 2 * (r - 1));
      if (rem > 0) {
        ans.push_back(v[i - 1] + r);
        m -= 1;
        rem -= 1;
        if (rem && m) {
          ans.push_back(v[i] - r);
          m -= 1;
        }
      }
    }
    if (m <= 0) break;
  }
  if (m > 0) ans.push_back(v[n - 1] + r);
  cout << p << '\n';
  for (auto c : ans) cout << c << " ";
  return 0;
}
