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>
#pragma optimization_level 3
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long long N = 3e+5 + 7;
const long long MAXN = 1e+9;
const long long mod = 998244353;
const long long INF = 0x7f7f7f7f7f7f7f7f;
const int INFi = 0x7f7f7f7f;
long long n, m, a[N];
vector<pair<long long, long long> > l, r;
vector<long long> ans;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  ;
  long long t;
  cin >> n >> m;
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a, a + n);
  for (int i = 0; i < n; i++) {
    if (i == 0) {
      l.push_back({1e9, a[i]});
    } else {
      l.push_back({ceil((a[i] - a[i - 1] - 1) / 2.0), a[i]});
    }
    if (i == 0) {
      r.push_back({1e9, a[n - 1]});
    } else {
      r.push_back({(a[i] - a[i - 1] - 1) / 2, a[i - 1]});
    }
  }
  sort(l.begin(), l.end());
  sort(r.begin(), r.end());
  long long l1 = 0, r1 = 0;
  long long ct = 1;
  long long res = 0;
  while (m > 0) {
    for (long long it = l1; it < n; it++) {
      if (m == 0) break;
      if (l[it].first <= 0)
        l1++;
      else {
        ans.push_back(l[it].second - ct);
        l[it].first--;
        res += ct;
        m--;
      }
    }
    for (long long it = r1; it < n; it++) {
      if (m == 0) break;
      if (r[it].first <= 0)
        r1++;
      else {
        ans.push_back(r[it].second + ct);
        r[it].first--;
        res += ct;
        m--;
      }
    }
    ct++;
  }
  cout << res << endl;
  for (auto it : ans) cout << it << " ";
}
