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;
constexpr long double m_pi = 3.1415926535897932L;
constexpr long long MOD = 1000000007;
constexpr long long INF = 1LL << 62;
constexpr long double EPS = 1e-10;
template <typename T>
using vector2 = vector<vector<T>>;
template <typename T>
using vector3 = vector<vector2<T>>;
string operator*(const string& s, int k) {
  if (k == 0) return "";
  string p = (s + s) * (k / 2);
  if (k % 2 == 1) p += s;
  return p;
}
template <class T>
inline bool chmin(T& a, T b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}
template <class T>
inline bool chmax(T& a, T b) {
  if (a < b) {
    a = b;
    return true;
  }
  return false;
}
struct Edge {
  int to, rev;
  long long cap;
  Edge(int _to, long long _cap, int _rev) {
    to = _to;
    cap = _cap;
    rev = _rev;
  }
};
void add_edge(vector<vector<Edge>>& G, int from, int to, long long cap,
              bool revFlag, long long revCap) {
  G[from].push_back(Edge(to, cap, (long long)G[to].size() + (from == to)));
  if (revFlag)
    G[to].push_back(Edge(from, revCap, (long long)G[from].size() - 1));
}
long long max_flow_dfs(vector<vector<Edge>>& G, long long v, long long t,
                       long long f, vector<bool>& used) {
  if (v == t) return f;
  used[v] = true;
  for (int i = 0; i < G[v].size(); ++i) {
    Edge& e = G[v][i];
    if (!used[e.to] && e.cap > 0) {
      long long d = max_flow_dfs(G, e.to, t, min(f, e.cap), used);
      if (d > 0) {
        e.cap -= d;
        G[e.to][e.rev].cap += d;
        return d;
      }
    }
  }
  return 0;
}
long long max_flow(vector<vector<Edge>>& G, long long s, long long t) {
  long long flow = 0;
  for (;;) {
    vector<bool> used(G.size());
    for (long long(i) = (long long)(0); i < (long long)(used.size()); i++)
      used[i] = false;
    long long f = max_flow_dfs(G, s, t, INF, used);
    if (f == 0) {
      return flow;
    }
    flow += f;
  }
}
void BellmanFord(vector<vector<Edge>>& G, long long s, vector<long long>& d,
                 vector<long long>& negative) {
  d.resize(G.size());
  negative.resize(G.size());
  for (long long(i) = (long long)(0); i < (long long)(d.size()); i++)
    d[i] = INF;
  for (long long(i) = (long long)(0); i < (long long)(d.size()); i++)
    negative[i] = false;
  d[s] = 0;
  for (long long(k) = (long long)(0); k < (long long)(G.size() - 1); k++) {
    for (long long(i) = (long long)(0); i < (long long)(G.size()); i++) {
      for (long long(j) = (long long)(0); j < (long long)(G[i].size()); j++) {
        if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
          d[G[i][j].to] = d[i] + G[i][j].cap;
        }
      }
    }
  }
  for (long long(k) = (long long)(0); k < (long long)(G.size() - 1); k++) {
    for (long long(i) = (long long)(0); i < (long long)(G.size()); i++) {
      for (long long(j) = (long long)(0); j < (long long)(G[i].size()); j++) {
        if (d[i] != INF && d[G[i][j].to] > d[i] + G[i][j].cap) {
          d[G[i][j].to] = d[i] + G[i][j].cap;
          negative[G[i][j].to] = true;
        }
        if (negative[i] == true) negative[G[i][j].to] = true;
      }
    }
  }
}
void Dijkstra(vector<vector<Edge>>& G, long long s, vector<long long>& d) {
  d.resize(G.size());
  for (long long(i) = (long long)(0); i < (long long)(d.size()); i++)
    d[i] = INF;
  d[s] = 0;
  priority_queue<pair<long long, long long>, vector<pair<long long, long long>>,
                 greater<pair<long long, long long>>>
      q;
  q.push(make_pair(0, s));
  while (!q.empty()) {
    pair<long long, long long> a = q.top();
    q.pop();
    if (d[a.second] < a.first) continue;
    for (long long(i) = (long long)(0); i < (long long)(G[a.second].size());
         i++) {
      Edge e = G[a.second][i];
      if (d[e.to] > d[a.second] + e.cap) {
        d[e.to] = d[a.second] + e.cap;
        q.push(make_pair(d[e.to], e.to));
      }
    }
  }
}
void WarshallFloyd(vector<vector<Edge>>& G, vector<vector<long long>>& d) {
  d.resize(G.size());
  for (long long(i) = (long long)(0); i < (long long)(d.size()); i++)
    d[i].resize(G.size());
  for (long long(i) = (long long)(0); i < (long long)(d.size()); i++) {
    for (long long(j) = (long long)(0); j < (long long)(d[i].size()); j++) {
      d[i][j] = ((i != j) ? INF : 0);
    }
  }
  for (long long(i) = (long long)(0); i < (long long)(G.size()); i++) {
    for (long long(j) = (long long)(0); j < (long long)(G[i].size()); j++) {
      chmin(d[i][G[i][j].to], G[i][j].cap);
    }
  }
  for (long long(i) = (long long)(0); i < (long long)(G.size()); i++) {
    for (long long(j) = (long long)(0); j < (long long)(G.size()); j++) {
      for (long long(k) = (long long)(0); k < (long long)(G.size()); k++) {
        chmin(d[j][k], d[j][i] + d[i][k]);
      }
    }
  }
}
bool tsort(vector<vector<Edge>>& graph, vector<long long>& order) {
  int n = graph.size(), k = 0;
  vector<long long> in(n);
  for (auto& es : graph)
    for (auto& e : es) in[e.to]++;
  priority_queue<long long, vector<long long>, greater<long long>> que;
  for (long long(i) = (long long)(0); i < (long long)(n); i++)
    if (in[i] == 0) que.push(i);
  while (que.size()) {
    int v = que.top();
    que.pop();
    order.push_back(v);
    for (auto& e : graph[v])
      if (--in[e.to] == 0) que.push(e.to);
  }
  if (order.size() != n)
    return false;
  else
    return true;
}
class Lca {
 public:
  const int n = 0;
  const int log2_n = 0;
  std::vector<std::vector<int>> parent;
  std::vector<int> depth;
  Lca() {}
  Lca(const vector<vector<Edge>>& g, int root)
      : n(g.size()),
        log2_n(log2(n) + 1),
        parent(log2_n, std::vector<int>(n)),
        depth(n) {
    dfs(g, root, -1, 0);
    for (int k = 0; k + 1 < log2_n; k++) {
      for (int v = 0; v < (int)g.size(); v++) {
        if (parent[k][v] < 0)
          parent[k + 1][v] = -1;
        else
          parent[k + 1][v] = parent[k][parent[k][v]];
      }
    }
  }
  void dfs(const vector<vector<Edge>>& g, int v, int p, int d) {
    parent[0][v] = p;
    depth[v] = d;
    for (auto& e : g[v]) {
      if (e.to != p) dfs(g, e.to, v, d + 1);
    }
  }
  int get(int u, int v) {
    if (depth[u] > depth[v]) std::swap(u, v);
    for (int k = 0; k < log2_n; k++) {
      if ((depth[v] - depth[u]) >> k & 1) {
        v = parent[k][v];
      }
    }
    if (u == v) return u;
    for (int k = log2_n - 1; k >= 0; k--) {
      if (parent[k][u] != parent[k][v]) {
        u = parent[k][u];
        v = parent[k][v];
      }
    }
    return parent[0][u];
  }
};
class UnionFind {
  vector<int> data;
  long long num;

 public:
  UnionFind(int size) : data(size, -1), num(size) {}
  bool unite(int x, int y) {
    x = root(x);
    y = root(y);
    if (x != y) {
      if (data[y] < data[x]) swap(x, y);
      data[x] += data[y];
      data[y] = x;
    }
    num -= (x != y);
    return x != y;
  }
  bool findSet(int x, int y) { return root(x) == root(y); }
  int root(int x) { return data[x] < 0 ? x : data[x] = root(data[x]); }
  long long size(int x) { return -data[root(x)]; }
  long long numSet() { return num; }
};
template <typename T, typename F>
class SegmentTree {
 private:
  T identity;
  F merge;
  long long n;
  vector<T> dat;

 public:
  SegmentTree(F f, T id, vector<T> v) : merge(f), identity(id) {
    int _n = v.size();
    n = 1;
    while (n < _n) n *= 2;
    dat.resize(2 * n - 1, identity);
    for (long long(i) = (long long)(0); i < (long long)(_n); i++)
      dat[n + i - 1] = v[i];
    for (int i = n - 2; i >= 0; i--)
      dat[i] = merge(dat[i * 2 + 1], dat[i * 2 + 2]);
  }
  SegmentTree(F f, T id, int _n) : merge(f), identity(id) {
    n = 1;
    while (n < _n) n *= 2;
    dat.resize(2 * n - 1, identity);
  }
  void set_val(int i, T x) {
    i += n - 1;
    dat[i] = x;
    while (i > 0) {
      i = (i - 1) / 2;
      dat[i] = merge(dat[i * 2 + 1], dat[i * 2 + 2]);
    }
  }
  T query(int l, int r) {
    T left = identity, right = identity;
    l += n - 1;
    r += n - 1;
    while (l < r) {
      if ((l & 1) == 0) left = merge(left, dat[l]);
      if ((r & 1) == 0) right = merge(dat[r - 1], right);
      l = l / 2;
      r = (r - 1) / 2;
    }
    return merge(left, right);
  }
};
template <typename T>
class FenwickTree {
  vector<T> data;
  int n;
  int p;

 public:
  FenwickTree(int n) : n(n) {
    data.resize(n + 1LL, 0);
    p = 1;
    while (p < data.size()) p *= 2;
  }
  T sum(int k) {
    T ret = 0;
    for (; k > 0; k -= k & -k) ret += data[k];
    return (ret);
  }
  T sum(int a, int b) { return sum(b) - sum(a); }
  void add(int k, T x) {
    for (++k; k <= n; k += k & -k) data[k] += x;
  }
  int lower_bound(long long w) {
    if (w <= 0) return -1;
    int x = 0;
    for (int k = p / 2; k > 0; k /= 2) {
      if (x + k <= n && data[x + k] < w) w -= data[x + k], x += k;
    }
    return x;
  }
};
void divisor(long long n, vector<long long>& ret) {
  for (long long i = 1; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back(i);
      if (i * i != n) ret.push_back(n / i);
    }
  }
  sort(ret.begin(), ret.end());
}
void prime_factorization(long long n, vector<pair<long long, long long>>& ret) {
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ret.push_back({i, 0});
      while (n % i == 0) {
        n /= i;
        ret[ret.size() - 1].second++;
      }
    }
  }
  if (n != 1) ret.push_back({n, 1});
}
inline long long mod_pow(long long x, long long n, long long mod) {
  long long res = 1;
  while (n > 0) {
    if (n & 1) res = res * x % mod;
    x = x * x % mod;
    n >>= 1;
  }
  return res;
}
inline long long mod_inv(long long x, long long mod) {
  return mod_pow(x, mod - 2, mod);
}
class Combination {
 public:
  vector<long long> fact;
  vector<long long> fact_inv;
  long long mod;
  long long nCr(long long n, long long r) {
    if (n < r) return 0;
    if (n < mod) return ((fact[n] * fact_inv[r] % mod) * fact_inv[n - r]) % mod;
    long long ret = 1;
    while (n || r) {
      long long _n = n % mod, _r = r % mod;
      n /= mod;
      r /= mod;
      (ret *= nCr(_n, _r)) %= mod;
    }
    return ret;
  }
  long long nPr(long long n, long long r) {
    return (fact[n] * fact_inv[n - r]) % mod;
  }
  long long nHr(long long n, long long r) { return nCr(r + n - 1, r); }
  Combination(long long _n, long long _mod) {
    mod = _mod;
    long long n = min(_n + 1, mod);
    fact.resize(n);
    fact[0] = 1;
    for (long long(i) = (long long)(0); i < (long long)(n - 1); i++) {
      fact[i + 1] = (fact[i] * (i + 1LL)) % mod;
    }
    fact_inv.resize(n);
    fact_inv[n - 1] = mod_inv(fact[n - 1], mod);
    for (int i = n - 1; i > 0; i--) {
      fact_inv[i - 1] = fact_inv[i] * i % mod;
    }
  }
};
long long popcount(long long x) {
  x = (x & 0x5555555555555555) + (x >> 1 & 0x5555555555555555);
  x = (x & 0x3333333333333333) + (x >> 2 & 0x3333333333333333);
  x = (x & 0x0F0F0F0F0F0F0F0F) + (x >> 4 & 0x0F0F0F0F0F0F0F0F);
  x = (x & 0x00FF00FF00FF00FF) + (x >> 8 & 0x00FF00FF00FF00FF);
  x = (x & 0x0000FFFF0000FFFF) + (x >> 16 & 0x0000FFFF0000FFFF);
  x = (x & 0x00000000FFFFFFFF) + (x >> 32 & 0x00000000FFFFFFFF);
  return x;
}
void solve() {
  long long n, p;
  cin >> n >> p;
  vector<long long> a(n);
  for (long long(i) = (long long)(0); i < (long long)(n); i++) cin >> a[i];
  sort((a).begin(), (a).end());
  long long mx = a.back();
  long long l = 0, r = mx + 1;
  while (r - l > 1) {
    long long m = (r + l) / 2;
    long long now = m;
    long long cnt = 0;
    long long left = 0;
    while (left != n && now >= a[left]) cnt++, left++;
    auto check = [&]() {
      for (long long(i) = (long long)(0); i < (long long)(n); i++) {
        if (left == n) return true;
        if (cnt == 0) return false;
        cnt--;
        now++;
        while (left != n && now >= a[left]) cnt++, left++;
      }
    };
    if (check() == true)
      r = m;
    else
      l = m;
  }
  long long s = 0;
  long long now = r;
  long long left = 0;
  while (left != n && now >= a[left]) s++, left++;
  long long cnt = s;
  map<long long, long long> st;
  vector<long long> x(n);
  for (long long(i) = (long long)(0); i < (long long)(n); i++) {
    x[i] = cnt;
    st[cnt]++;
    cnt--;
    now++;
    while (left != n && now >= a[left]) cnt++, left++;
  }
  map<long long, long long> mp;
  for (long long(i) = (long long)(0); i < (long long)(n); i++) mp[a[i]]++;
  vector<long long> ans;
  for (long long(i) = (long long)(0); i < (long long)(n); i++) {
    if (s >= p) break;
    st[x[i]]--;
    if (st[x[i]] == 0) st.erase(x[i]);
    if (st.rbegin()->first + i >= p) break;
    ans.push_back(r + i);
    now++;
    s += mp[r + i + 1];
  }
  cout << ans.size() << "\n";
  for (long long(i) = (long long)(0); i < (long long)(ans.size()); i++)
    cout << ans[i] << " ";
  cout << "\n";
}
int main() {
  ios::sync_with_stdio(false);
  std::cin.tie(0);
  std::cout.tie(0);
  solve();
  return 0;
}