Problem: This is the easy version of the problem. The difference between versions is the constraints on n and a_i. You can make hacks only if all versions of the problem are solved.

First, Aoi came up with the following idea for the competitive programming problem:

Yuzu is a girl who collecting candies. Originally, she has x candies. There are also n enemies numbered with integers from 1 to n. Enemy i has a_i candies.

Yuzu is going to determine a permutation P. A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, \{2,3,1,5,4\} is a permutation, but \{1,2,2\} is not a permutation (2 appears twice in the array) and \{1,3,4\} is also not a permutation (because n=3 but there is the number 4 in the array).

After that, she will do n duels with the enemies with the following rules:

  * If Yuzu has equal or more number of candies than enemy P_i, she wins the duel and gets 1 candy. Otherwise, she loses the duel and gets nothing. 
  * The candy which Yuzu gets will be used in the next duels. 



Yuzu wants to win all duels. How many valid permutations P exist?

This problem was easy and wasn't interesting for Akari, who is a friend of Aoi. And Akari made the following problem from the above idea:

Let's define f(x) as the number of valid permutations for the integer x.

You are given n, a and a prime number p ≤ n. Let's call a positive integer x good, if the value f(x) is not divisible by p. Find all good integers x.

Your task is to solve this problem made by Akari.

Input

The first line contains two integers n, p (2 ≤ p ≤ n ≤ 2000). It is guaranteed, that the number p is prime (it has exactly two divisors 1 and p).

The second line contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 2000).

Output

In the first line, print the number of good integers x.

In the second line, output all good integers x in the ascending order.

It is guaranteed that the number of good integers x does not exceed 10^5.

Examples

Input


3 2
3 4 5


Output


1
3


Input


4 3
2 3 5 6


Output


2
3 4


Input


4 3
9 1 1 1


Output


0

Note

In the first test, p=2.

  * If x ≤ 2, there are no valid permutations for Yuzu. So f(x)=0 for all x ≤ 2. The number 0 is divisible by 2, so all integers x ≤ 2 are not good. 
  * If x = 3, \{1,2,3\} is the only valid permutation for Yuzu. So f(3)=1, so the number 3 is good. 
  * If x = 4, \{1,2,3\} , \{1,3,2\} , \{2,1,3\} , \{2,3,1\} are all valid permutations for Yuzu. So f(4)=4, so the number 4 is not good. 
  * If x ≥ 5, all 6 permutations are valid for Yuzu. So f(x)=6 for all x ≥ 5, so all integers x ≥ 5 are not good. 



So, the only good number is 3.

In the third test, for all positive integers x the value f(x) is divisible by p = 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
using ll = int64_t;
using ld = long double;
using P = pair<int, int>;
using vs = vector<string>;
using vi = vector<int>;
using vvi = vector<vi>;
template <class T>
using PQ = priority_queue<T>;
template <class T>
using PQG = priority_queue<T, vector<T>, greater<T> >;
const int INF = 100010001;
const ll LINF = (ll)INF * INF * 10;
template <typename T1, typename T2>
inline bool chmax(T1 &a, T2 b) {
  return a < b && (a = b, true);
}
template <typename T1, typename T2>
inline bool chmin(T1 &a, T2 b) {
  return a > b && (a = b, true);
}
template <typename T1, typename T2>
istream &operator>>(istream &is, pair<T1, T2> &p) {
  return is >> p.first >> p.second;
}
template <typename T1, typename T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &p) {
  return os << p.first << ' ' << p.second;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, p;
  cin >> n >> p;
  vi a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  sort(a.begin(), a.end());
  int ue = a[n - 1], sita = 0;
  bool dame[4001] = {};
  int cnt[2001] = {};
  int j = 0;
  for (int i = 1; i <= a[n - 1]; i++) {
    cnt[i] = cnt[i - 1];
    while (j < n and a[j] == i) {
      j++;
      cnt[i]++;
    }
    chmax(sita, i - cnt[i]);
    for (int k = 0; k < min(i, cnt[i]); ++k) {
      if ((cnt[i] - k) % p == 0) {
        dame[i - k] = true;
      }
    }
  }
  for (int i = a[n - 1] + 1; i <= a[n - 1] + p; i++) {
    for (int k = n % p; k < min(n, i); k += p) {
      dame[i - k] = true;
    }
  }
  vi ans;
  for (int i = sita + 1; i < ue; i++)
    if (not dame[i]) {
      ans.push_back(i);
    }
  cout << ans.size() << endl;
  for (int i = 0; i < ans.size(); ++i) {
    cout << ans[i] << (i == ans.size() - 1 ? '\n' : ' ');
  }
}