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 = 2e5 + 10;
struct node {
  int l, r, sum;
} p[maxn];
int a[maxn];
int ans[maxn];
bool cmp(node xx, node yy) { return xx.sum < yy.sum; }
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + 1 + n);
  if (n == 1) {
    int pre = 0, cnt = 1;
    long long sum = 0;
    while (cnt <= m) {
      pre++;
      ans[cnt++] = a[1] - pre;
      sum += pre;
      if (cnt > m) break;
      ans[cnt++] = a[1] + pre;
      sum += pre;
    }
    cout << sum << endl;
    for (int i = 1; i <= m; i++) cout << ans[i] << " ";
    return 0;
  }
  int num = 1;
  for (int i = 2; i <= n; i++) {
    if (a[i] - a[i - 1] == 1) continue;
    p[num].l = a[i - 1] + 1;
    p[num].r = a[i] - 1;
    p[num].sum = p[num].r - p[num].l + 1;
    num++;
  }
  sort(p + 1, p + num, cmp);
  int cont = 1, cnt = 1, pre = 1, flag = 1;
  int p1 = a[1] - 1;
  int p2 = a[n] + 1;
  long long qum = 0;
  while (cnt <= m) {
    for (int i = cont; cnt <= m && i < num; i++) {
      ans[cnt++] = p[i].l++;
      qum += pre;
      p[i].sum--;
      if (cnt > m || p[i].sum == 0) {
        cont++;
        continue;
      }
      ans[cnt++] = p[i].r--;
      p[i].sum--;
      qum += pre;
      if (p[i].sum == 0) cont++;
    }
    if (cnt <= m) {
      ans[cnt++] = p1--;
      qum += pre;
    }
    if (cnt <= m) {
      ans[cnt++] = p2++;
      qum += pre;
    }
    pre++;
  }
  cout << qum << endl;
  for (int i = 1; i <= m; i++) cout << ans[i] << " ";
}
