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;
  mt19937 rnd(time(0));
  struct Node {
    int val, pos;
    bool operator<(const Node& ot) const { return val < ot.val; }
  };
  struct lazy_segment_tree {
    static const int N = 1e6 + 10;
    const Node good_value = {0, 0};
    Node t[4 * N], ch[4 * N];
    Node merge(Node a, Node b) { return max(a, b); }
    void build(int v = 1, int tl = 1, int tr = N - 1) {
      if (tl == tr) {
        t[v] = {0, 0};
        return;
      }
      int tm = (tl + tr) / 2;
      build(v + v, tl, tm);
      build(v + v + 1, tm + 1, tr);
      t[v] = merge(t[v + v], t[v + v + 1]);
    }
    void push(int v, int tl, int tr) {
      if (!ch[v].val) return;
      t[v] = ch[v];
      if (tl < tr) {
        ch[v + v] = ch[v];
        ch[v + v + 1] = ch[v];
      }
      ch[v] = {0, 0};
    }
    void update(int l, int r, Node x, int v = 1, int tl = 1, int tr = N - 1) {
      push(v, tl, tr);
      if (r < tl || tr < l) return;
      if (l <= tl && tr <= r) {
        ch[v] = x;
        push(v, tl, tr);
        return;
      }
      int tm = (tl + tr) / 2;
      update(l, r, x, v + v, tl, tm);
      update(l, r, x, v + v + 1, tm + 1, tr);
      t[v] = merge(t[v + v], t[v + v + 1]);
    }
    Node get(int l, int r, int v = 1, int tl = 1, int tr = N - 1) {
      push(v, tl, tr);
      if (r < tl || tr < l) return good_value;
      if (l <= tl && tr <= r) return t[v];
      int tm = (tl + tr) / 2;
      Node res =
          merge(get(l, r, v + v, tl, tm), get(l, r, v + v + 1, tm + 1, tr));
      t[v] = merge(t[v + v], t[v + v + 1]);
      return res;
    }
  } tree;
  const int N = 3e5 + 10;
  vector<pair<int, int> > a[N];
  vector<int> all;
  map<int, int> mp;
  int mem[N];
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= m; i++) {
      int id, l, r;
      cin >> id >> l >> r;
      a[id].push_back({l, r});
      all.push_back(l);
      all.push_back(r);
    }
    sort(all.begin(), all.end());
    all.erase(unique(all.begin(), all.end()), all.end());
    for (int i = 0; i < all.size(); i++) {
      mp[all[i]] = i + 1;
    }
    tree.build();
    for (int i = 1; i <= n; i++) {
      int cur = 0;
      for (auto& [l, r] : a[i]) {
        l = mp[l];
        r = mp[r];
        Node tmp = tree.get(l, r);
        tmp.val++;
        if (cur < tmp.val) {
          cur = tmp.val;
          mem[i] = tmp.pos;
        }
      }
      for (int j = 0; j < a[i].size(); j++) {
        tree.update(a[i][j].first, a[i][j].second, {cur, i});
      }
    }
    Node res = tree.get(1, tree.N - 1);
    cout << n - res.val << "\n";
    vector<int> ans;
    int ptr = res.pos;
    for (int i = n; i >= ptr + 1; i--) {
      ans.push_back(i);
    }
    while (ptr != mem[ptr]) {
      for (int i = ptr - 1; i >= mem[ptr] + 1; i--) {
        ans.push_back(i);
      }
      ptr = mem[ptr];
    }
    reverse(ans.begin(), ans.end());
    for (int i : ans) {
      cout << i << " ";
    }
    cout << "\n";
    return 0;
  }
  