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;
  const int N = 300300;
  vector<pair<int, int> > segs[N];
  struct node {
    int l, r, mid;
    pair<int, int> mx;
    node *L, *R;
    bool prop;
    node(int ll, int rr) {
      l = ll, r = rr;
      mid = (l + r) / 2;
      mx = {0, -1};
      prop = 0;
      if (l == r) return;
      L = new node(l, mid);
      R = new node(mid + 1, r);
    }
    pair<int, int> query(int x, int y) {
      propagate_max();
      if (x == l && y == r) return mx;
      if (y <= mid) return L->query(x, y);
      if (x > mid) return R->query(x, y);
      return max(L->query(x, mid), R->query(mid + 1, y));
    }
    void propagate_max() {
      if (prop) {
        if (l != r) {
          L->mx = R->mx = mx;
          L->prop = R->prop = true;
        }
        prop = 0;
      }
    }
    void set_max(int x, int y, int n_mx, int n_nx) {
      propagate_max();
      if (x == l && y == r) {
        mx = {n_mx, n_nx};
        prop = true;
        return;
      }
      if (y <= mid)
        L->set_max(x, y, n_mx, n_nx);
      else if (x > mid)
        R->set_max(x, y, n_mx, n_nx);
      else {
        L->set_max(x, mid, n_mx, n_nx);
        R->set_max(mid + 1, y, n_mx, n_nx);
      }
      mx = max(L->mx, R->mx);
    }
  };
  int main() {
    int n, m;
    scanf("%d %d", &n, &m);
    vector<int> zip;
    for (int i = 0; i < m; ++i) {
      int row, l, r;
      scanf("%d %d %d", &row, &l, &r);
      --row;
      segs[row].push_back({l, r});
      zip.push_back(l);
      zip.push_back(r);
    }
    sort(zip.begin(), zip.end());
    zip.resize(unique(zip.begin(), zip.end()) - zip.begin());
    m = (int)zip.size();
    vector<pair<int, int> > ans(n);
    int best = n - 1;
    node st(0, m - 1);
    for (int i = n - 1; i >= 0; --i) {
      ans[i] = {0, -1};
      for (auto sg : segs[i]) {
        int idx_l = lower_bound(zip.begin(), zip.end(), sg.first) - zip.begin();
        int idx_r = lower_bound(zip.begin(), zip.end(), sg.second) - zip.begin();
        auto cand = st.query(idx_l, idx_r);
        ans[i] = max(ans[i], cand);
      }
      ++ans[i].first;
      if (ans[i] > ans[best]) best = i;
      for (auto sg : segs[i]) {
        int idx_l = lower_bound(zip.begin(), zip.end(), sg.first) - zip.begin();
        int idx_r = lower_bound(zip.begin(), zip.end(), sg.second) - zip.begin();
        st.set_max(idx_l, idx_r, ans[i].first, i);
      }
    }
    vector<bool> good(n);
    while (best != -1) {
      good[best] = 1;
      best = ans[best].second;
    }
    vector<int> bad;
    for (int i = 0; i < n; ++i)
      if (!good[i]) bad.push_back(i);
    int sz = (int)bad.size();
    printf("%d\n", sz);
    for (int i = 0; i < sz; ++i) printf("%d%c", ++bad[i], " \n"[i == sz - 1]);
    return 0;
  }
  