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 GCC optimize(3)
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-pointer", "inline")
using namespace std;
const long long INF = 1e16;
const int maxn = 2e5 + 6;
const int mod = 1e9 + 6;
const double eps = 1e-15;
inline bool read(long long &num) {
  char in;
  bool IsN = false;
  in = getchar();
  if (in == EOF) return false;
  while (in != '-' && (in < '0' || in > '9')) in = getchar();
  if (in == '-') {
    IsN = true;
    num = 0;
  } else
    num = in - '0';
  while (in = getchar(), in >= '0' && in <= '9') {
    num *= 10, num += in - '0';
  }
  if (IsN) num = -num;
  return true;
}
long long n, m, p;
long long num[maxn];
int cnt = 0;
int b[maxn];
map<int, int> mp;
long long bfs() {
  long long ans = 0;
  queue<pair<int, int> > q;
  for (int i = 1; i <= n; i++) {
    q.push({num[i], num[i]});
    mp[num[i]] = 1;
  }
  while (!q.empty()) {
    auto p = q.front();
    q.pop();
    if (!mp[p.second + 1]) {
      mp[p.second + 1] = 1;
      ans += abs(p.second + 1 - p.first);
      b[++cnt] = p.second + 1;
      mp[p.second + 1] = 1;
      if (!mp[p.second + 2]) q.push({p.first, p.second + 1});
      if (cnt == m) return ans;
    }
    if (!mp[p.second - 1]) {
      mp[p.second - 1] = 1;
      ans += abs(p.second - 1 - p.first);
      b[++cnt] = p.second - 1;
      mp[p.second - 1] = 1;
      if (!mp[p.second - 2]) q.push({p.first, p.second - 1});
      if (cnt == m) return ans;
    }
  }
}
int main() {
  read(n);
  read(m);
  for (int i = 1; i <= n; i++) read(num[i]);
  printf("%lld\n", bfs());
  for (int i = 1; i <= m; i++) printf("%d ", b[i]);
  return 0;
}
