description: Moamen was drawing a grid of n rows and 10^9 columns containing only digits 0 and 1. Ezzat noticed what Moamen was drawing and became interested in the minimum number of rows one needs to remove to make the grid beautiful.
  
  A grid is beautiful if and only if for every two consecutive rows there is at least one column containing 1 in these two rows.
  
  Ezzat will give you the number of rows n, and m segments of the grid that contain digits 1. Every segment is represented with three integers i, l, and r, where i represents the row number, and l and r represent the first and the last column of the segment in that row.
  
  For example, if n = 3, m = 6, and the segments are (1,1,1), (1,7,8), (2,7,7), (2,15,15), (3,1,1), (3,15,15), then the grid is:
  
  <image>
  
  Your task is to tell Ezzat the minimum number of rows that should be removed to make the grid beautiful.
  
  Input
  
  The first line contains two integers n and m (1 ≤ n, m ≤ 3⋅10^5).
  
  Each of the next m lines contains three integers i, l, and r (1 ≤ i ≤ n, 1 ≤ l ≤ r ≤ 10^9). Each of these m lines means that row number i contains digits 1 in columns from l to r, inclusive.
  
  Note that the segments may overlap.
  
  Output
  
  In the first line, print a single integer k — the minimum number of rows that should be removed.
  
  In the second line print k distinct integers r_1, r_2, …, r_k, representing the rows that should be removed (1 ≤ r_i ≤ n), in any order.
  
  If there are multiple answers, print any.
  
  Examples
  
  Input
  
  
  3 6
  1 1 1
  1 7 8
  2 7 7
  2 15 15
  3 1 1
  3 15 15
  
  
  Output
  
  
  0
  
  
  Input
  
  
  5 4
  1 2 3
  2 4 6
  3 3 5
  5 1 1
  
  
  Output
  
  
  3
  2 4 5
  
  Note
  
  In the first test case, the grid is the one explained in the problem statement. The grid has the following properties: 
  
    1. The 1-st row and the 2-nd row have a common 1 in the column 7. 
    2. The 2-nd row and the 3-rd row have a common 1 in the column 15. 
  
  As a result, this grid is beautiful and we do not need to remove any row.
  
  In the second test case, the given grid is as follows: 
  
  <image>
solution: #include <bits/stdc++.h>
  using namespace std;
  using db = double;
  using s64 = long long;
  using ld = long double;
  using u32 = unsigned int;
  using u64 = unsigned long long;
  using pii = pair<int, int>;
  const double PI = acos(-1);
  const double eps = 1e-7;
  namespace Solver {
  const char endl = '\n';
  const int MN = 3e5 + 5;
  int n, m, dp[MN], pre[MN];
  inline void ckmin(int& x, int y) {
    if (dp[y] < dp[x]) x = y;
  }
  namespace SegT {
  struct Node {
    Node *ls, *rs;
    int tag, val;
    Node() {
      ls = rs = nullptr;
      tag = val = 0;
    }
  }* rt = nullptr;
  void add(int l, int r, int v, Node*& o = rt, int L = 1, int R = 1e9) {
    if (r < L || R < l) return;
    if (o == nullptr) o = new Node;
    if (l <= L && R <= r) {
      ckmin(o->tag, v);
      ckmin(o->val, v);
      return;
    }
    int mid = L + R >> 1;
    add(l, r, v, o->ls, L, mid);
    add(l, r, v, o->rs, mid + 1, R);
    if (o->ls != nullptr) ckmin(o->val, o->ls->val);
    if (o->rs != nullptr) ckmin(o->val, o->rs->val);
  }
  int query(int l, int r, Node* o = rt, int L = 1, int R = 1e9) {
    if (o == nullptr || r < L || R < l) return 0;
    if (l <= L && R <= r) return o->val;
    int mid = L + R >> 1, res = o->tag;
    ckmin(res, query(l, r, o->ls, L, mid));
    ckmin(res, query(l, r, o->rs, mid + 1, R));
    return res;
  }
  }  // namespace SegT
  struct O {
    int i, l, r;
  } o[MN];
  void main() {
    cin >> n >> m;
    for (decay<decltype(m)>::type i = (0), _i = (m); i < _i; ++i)
      cin >> o[i].i >> o[i].l >> o[i].r;
    sort(o, o + m, [](const O& a, const O& b) { return a.i < b.i; });
    for (int i = 0, j; i < m; i = j) {
      int I = o[i].i;
      for (j = i; o[j].i == I; ++j) ckmin(pre[I], SegT::query(o[j].l, o[j].r));
      dp[I] = dp[pre[I]] - 1;
      for (decay<decltype(j)>::type k = (i), _k = (j); k < _k; ++k)
        SegT::add(o[k].l, o[k].r, I);
    }
    int ans = n, p = 0;
    for (decay<decltype(n)>::type i = (1), _i = (n); i <= _i; ++i)
      if (dp[i] + n < ans) ans = dp[i] + n, p = i;
    cout << ans << endl;
    vector<bool> used(n + 1, false);
    for (int i = p; i; i = pre[i]) used[i] = true;
    for (decay<decltype(n)>::type i = (1), _i = (n); i <= _i; ++i)
      if (!used[i]) cout << i << ' ';
    cout << endl;
  }
  }  // namespace Solver
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int T = 1;
    for (decay<decltype(T)>::type _t = (1), __t = (T); _t <= __t; ++_t) {
      Solver::main();
    }
    return 0;
  }
  