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>
  struct Node {
    int l = 0, r = 0;
    std::pair<int, int> v = {0, 0}, p = {0, 0};
  } static_tree[(int)1e7];
  struct SegTree {
    Node *t = static_tree;
    int sz = 2;
    SegTree() { t[0] = t[1] = Node(); }
    inline void visit(int n, int l, int m, int r) {
      if (l <= m) {
        if (!t[n].l) t[n].l = sz++, t[t[n].l] = Node();
        t[t[n].l].v = std::max(t[t[n].l].v, t[n].p);
        t[t[n].l].p = std::max(t[t[n].l].p, t[n].p);
      }
      if (m + 1 <= r) {
        if (!t[n].r) t[n].r = sz++, t[t[n].r] = Node();
        t[t[n].r].v = std::max(t[t[n].r].v, t[n].p);
        t[t[n].r].p = std::max(t[t[n].r].p, t[n].p);
      }
      t[n].p = {0, 0};
    }
    std::pair<int, int> query(int n, int l, int r, int lq, int rq) {
      if (r < lq || rq < l)
        return {0, 0};
      else if (lq <= l && r <= rq)
        return t[n].v;
      int m = l + (r - l) / 2;
      visit(n, l, m, r);
      return std::max(query(t[n].l, l, m, lq, rq),
                      query(t[n].r, m + 1, r, lq, rq));
    }
    void update(int n, int l, int r, int lq, int rq, std::pair<int, int> mx) {
      if (r < lq || rq < l) {
        return;
      } else if (lq <= l && r <= rq) {
        t[n].v = std::max(t[n].v, mx);
        t[n].p = std::max(t[n].p, mx);
      } else {
        int m = l + (r - l) / 2;
        visit(n, l, m, r);
        update(t[n].l, l, m, lq, rq, mx);
        update(t[n].r, m + 1, r, lq, rq, mx);
        t[n].v = std::max(t[t[n].l].v, t[t[n].r].v);
      }
    }
  };
  int bptr[(int)3e5 + 10];
  std::vector<std::pair<int, int>> intervals[(int)3e5 + 10];
  int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    for (int x = 0; x < m; ++x) {
      int i, l, r;
      scanf("%d %d %d", &i, &l, &r);
      intervals[i].push_back({l, r});
    }
    SegTree tree;
    for (int i = 1; i <= n; ++i) {
      if (!intervals[i].size()) continue;
      auto &v = intervals[i];
      std::vector<std::pair<int, int>> merged;
      std::sort(v.begin(), v.end(), std::greater<std::pair<int, int>>{});
      while (!v.empty()) {
        int l = v.back().first, r = v.back().second;
        v.pop_back();
        while (!v.empty() && v.back().first <= r) {
          r = std::max(r, v.back().second);
          v.pop_back();
        }
        merged.push_back({l, r});
      }
      std::pair<int, int> dp = {0, 0};
      for (auto &lr : merged) {
        int l = lr.first, r = lr.second;
        dp = std::max(dp, tree.query(1, 1, 1e9 + 10, l, r));
      }
      for (auto &lr : merged) {
        int l = lr.first, r = lr.second;
        tree.update(1, 1, 1e9 + 10, l, r, {dp.first + 1, i});
      }
      bptr[i] = dp.second;
    }
    std::unordered_set<int> ans;
    for (int i = tree.query(1, 1, 1e9 + 10, 1, 1e9 + 10).second; i; i = bptr[i])
      ans.insert(i);
    printf("%d\n", n - ans.size());
    if (ans.size() != n) {
      for (int i = 1; i <= n; ++i)
        if (!ans.count(i)) printf("%d ", i);
      printf("\n");
    }
  }
  