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 SegTree {
    struct Node {
      int l = 0, r = 0;
      std::pair<int, int> v = {0, -1}, p = {0, -1};
    };
    std::vector<Node> tree = {{}, {}};
    void visit(int n, int l, int m, int r) {
      if (!tree[n].l && l <= m) tree[n].l = tree.size(), tree.emplace_back();
      if (!tree[n].r && m + 1 <= r) tree[n].r = tree.size(), tree.emplace_back();
      setMax(tree[n].l, l, m, l, r, tree[n].p);
      setMax(tree[n].r, m + 1, r, l, r, tree[n].p);
      tree[n].p = Node().p;
    }
    void setMax(int n, int l, int r, int lq, int rq, std::pair<int, int> v) {
      if (!n || r < lq || rq < l || rq < lq) {
        return;
      } else if (lq <= l && r <= rq) {
        tree[n].v = std::max(tree[n].v, v);
        tree[n].p = std::max(tree[n].p, v);
      } else {
        int m = l + (r - l) / 2;
        visit(n, l, m, r);
        setMax(tree[n].l, l, m, lq, rq, v);
        setMax(tree[n].r, m + 1, r, lq, rq, v);
        tree[n].v = std::max(tree[tree[n].l].v, tree[tree[n].r].v);
      }
    }
    std::pair<int, int> getMax(int n, int l, int r, int lq, int rq) {
      if (!n || r < lq || rq < l || rq < lq)
        return Node().v;
      else if (lq <= l && r <= rq)
        return tree[n].v;
      int m = l + (r - l) / 2;
      visit(n, l, m, r);
      return std::max(getMax(tree[n].l, l, m, lq, rq),
                      getMax(tree[n].r, m + 1, r, lq, rq));
    }
  } tree;
  int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    std::vector<int> points;
    std::unordered_map<int, int> relpoint;
    std::vector<std::vector<std::pair<int, int>>> intervals(n + 1);
    for (int x = 0, i, l, r; x < m; ++x) {
      scanf("%d %d %d", &i, &l, &r);
      intervals[i].push_back({l, r});
      if (!relpoint.count(l)) points.push_back(l);
      if (!relpoint.count(r)) points.push_back(r);
      relpoint[l] = relpoint[r] = 0;
    }
    std::sort(points.begin(), points.end());
    for (int i = 1; i <= points.size(); ++i) relpoint[points[i]] = i;
    std::vector<int> bptr(n + 1, -1);
    for (int row = 1; row <= n; ++row) {
      auto &v = intervals[row];
      std::sort(v.begin(), v.end());
      std::vector<std::pair<int, int>> ranges;
      for (int i = 0, j; i < v.size(); i = j) {
        int l = v[i].first, r = v[i].second;
        for (j = i + 1; j < v.size() && v[j].first <= r; ++j)
          r = std::max(r, v[j].second);
        ranges.push_back({relpoint[l], relpoint[r]});
      }
      std::pair<int, int> dp = {0, -1};
      for (auto &lr : ranges)
        dp = std::max(dp, tree.getMax(1, 1, points.size(), lr.first, lr.second));
      for (auto &lr : ranges) {
        tree.setMax(1, 1, points.size(), lr.first, lr.second,
                    {dp.first + 1, row});
      }
      bptr[row] = dp.second;
    }
    auto ans = tree.getMax(1, 1, points.size(), 1, points.size());
    printf("%d\n", n - ans.first);
    for (int i = ans.second, j; i != -1;) i = bptr[j = i], bptr[j] = 0;
    for (int row = 1; row <= n; ++row)
      if (bptr[row] != 0) printf("%d ", row);
    if (ans.first != n) printf("\n");
  }
  