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 int MAXN = 1000002;
const int MOD2 = 998244353;
const int MOD = 1000000007;
const int INF = 1e8;
const long double EPS = 1e-7;
long long int mul(long long int x, long long int y) {
  return (x * 1ll * y) % MOD;
}
long long int fastpow(long long int x, long long int y) {
  long long int z = 1;
  while (y) {
    if (y & 1) z = mul(z, x);
    x = mul(x, x);
    y >>= 1;
  }
  return z;
}
long long int modinv(long long int n, long long int p) {
  return fastpow(n, p - 2);
}
struct Comp {
  bool operator()(const std::pair<int, int>& a, const std::pair<int, int>& b) {
    if (a.first != b.first) {
      return a.first > b.first;
    }
    return a.second > b.second;
  }
};
int main(int argc, char** argv) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  {
    long long int n, m;
    cin >> n >> m;
    long long int a[n];
    for (int i = 0; i < (int)n; i++) cin >> a[i];
    sort(a, a + n);
    map<long long int, long long int> vis, vis2;
    for (int i = 0; i < (int)n; i++) {
      vis[a[i]]++;
      vis2[a[i]]++;
    }
    priority_queue<pair<long long int, long long int>> pq;
    for (int i = 0; i < (int)n; i++) {
      if (i > 0 && a[i] - a[i - 1] <= 2) {
        if (vis[a[i] + 1] == 0) {
          pq.push({-1, a[i] + 1});
        }
        continue;
      }
      if (vis[a[i] - 1] == 0) {
        pq.push({-1, a[i] - 1});
        vis[a[i] - 1]++;
      }
      if (vis[a[i] + 1] == 0) {
        pq.push({-1, a[i] + 1});
      }
    }
    long long int cnt = 0, ansm = 0;
    vector<long long int> ans;
    while (cnt < m && !pq.empty()) {
      long long int x = pq.top().first;
      long long int y = pq.top().second;
      pq.pop();
      if (vis2[y]) {
        if (vis[y + 1] == 0) {
          pq.push({x - 1, y + 1});
          vis[y + 1]++;
        }
        if (vis[y - 1] == 0) {
          pq.push({x - 1, y - 1});
          vis[y - 1]++;
        }
        continue;
      }
      vis2[y] = 1;
      ansm -= x;
      vis[y]++;
      ans.push_back(y);
      if (vis[y + 1] == 0) {
        pq.push({x - 1, y + 1});
        vis[y + 1]++;
      }
      if (vis[y - 1] == 0) {
        pq.push({x - 1, y - 1});
        vis[y - 1]++;
      }
      cnt++;
    }
    cout << ansm << "\n";
    for (int i = 0; i < (int)cnt; i++) cout << ans[i] << " ";
  }
  return 0;
}
