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;
  struct Node {
    Node* left = nullptr;
    Node* right = nullptr;
    pair<int, int> mmax = {0, 0};
    pair<int, int> assign = {-1, -1};
    Node() {}
    Node(pair<int, int> mm, pair<int, int> a)
        : left(nullptr), right(nullptr), mmax(mm), assign(a) {}
    void push() {
      if (assign.second != -1) {
        left->assign = assign;
        right->assign = assign;
      }
      assign = {-1, -1};
    }
    void create_children() {
      if (!left) {
        left = new Node();
      }
      if (!right) {
        right = new Node();
      }
    }
    void update() {
      mmax = max((left->assign.second != -1 ? left->assign : left->mmax),
                 (right->assign.second != -1 ? right->assign : right->mmax));
    }
    pair<int, int> qmax(int lq, int rq, int lb = 0, int rb = 7e5 + 1) {
      if (lq <= lb && rb <= rq) {
        return (assign.second != -1 ? assign : mmax);
      } else if (lq >= rb || rq <= lb) {
        return {0, 0};
      }
      create_children();
      push();
      int mid = (lb + rb) >> 1;
      auto res = max(left->qmax(lq, rq, lb, mid), right->qmax(lq, rq, mid, rb));
      update();
      return res;
    }
    void qassign(int lq, int rq, pair<int, int> val, int lb = 0,
                 int rb = 7e5 + 1) {
      if (lq <= lb && rb <= rq) {
        assign = val;
        return;
      } else if (lq >= rb || rq <= lb) {
        return;
      }
      create_children();
      push();
      int mid = (lb + rb) >> 1;
      left->qassign(lq, rq, val, lb, mid);
      right->qassign(lq, rq, val, mid, rb);
      update();
    }
    ~Node() {
      delete left;
      delete right;
    }
  };
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, m;
    cin >> n >> m;
    vector<vector<pair<int, int>>> ranges(1 + n);
    vector<int> coord;
    for (int i = 0; i < m; ++i) {
      int ind, left, right;
      cin >> ind >> left >> right;
      ++right;
      ranges[ind].push_back({left, right});
      coord.push_back(left);
      coord.push_back(right);
    }
    sort(coord.begin(), coord.end());
    coord.resize(unique(coord.begin(), coord.end()) - coord.begin());
    for (int i = 1; i <= n; ++i) {
      for (pair<int, int>& r : ranges[i]) {
        r.first =
            lower_bound(coord.begin(), coord.end(), r.first) - coord.begin();
        r.second =
            lower_bound(coord.begin(), coord.end(), r.second) - coord.begin();
      }
    }
    vector<int> dp(1 + n, 0);
    vector<int> prev(1 + n, -1);
    Node* root = new Node();
    for (int i = 1; i <= n; ++i) {
      pair<int, int> mm = {0, 0};
      for (pair<int, int> rr : ranges[i]) {
        mm = max(mm, root->qmax(rr.first, rr.second));
      }
      dp[i] = mm.first + 1;
      prev[i] = mm.second;
      for (pair<int, int> rr : ranges[i]) {
        root->qassign(rr.first, rr.second, {mm.first + 1, i});
      }
    }
    int ind = max_element(dp.begin(), dp.end()) - dp.begin();
    cout << n - dp[ind] << "\n";
    vector<bool> need(1 + n, false);
    while (ind != 0) {
      need[ind] = true;
      ind = prev[ind];
    }
    for (int i = 1; i <= n; ++i) {
      if (!need[i]) {
        cout << i << " ";
      }
    }
  }
  