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 maxn = 312345;
  struct Seg {
    int l, r;
    bool lzy;
    pair<int, int> v;
  } T[maxn << 3];
  vector<pair<int, int> > G[maxn];
  int nxt[maxn];
  bool ans[maxn];
  void build(int o, int l, int r) {
    T[o].l = l, T[o].r = r;
    T[o].v = {0, -1};
    if (l < r) {
      int mi = l + r >> 1;
      build(o << 1, l, mi);
      build(o << 1 | 1, mi + 1, r);
    }
  }
  void push_down(int o) {
    if (T[o].lzy) {
      T[o << 1].v = T[o << 1 | 1].v = T[o].v;
      T[o << 1].lzy = T[o << 1 | 1].lzy = true;
      T[o].lzy = false;
    }
  }
  pair<int, int> query(int o, int l, int r) {
    if (l <= T[o].l && T[o].r <= r) return T[o].v;
    push_down(o);
    int mi = T[o].l + T[o].r >> 1;
    pair<int, int> ret = {0, -1};
    if (l <= mi) ret = max(ret, query(o << 1, l, r));
    if (r > mi) ret = max(ret, query(o << 1 | 1, l, r));
    return ret;
  }
  void update(int o, int l, int r, pair<int, int> v) {
    if (l <= T[o].l && T[o].r <= r) {
      T[o].lzy = true;
      T[o].v = v;
      return;
    }
    int mi = T[o].l + T[o].r >> 1;
    if (l <= mi) update(o << 1, l, r, v);
    if (r > mi) update(o << 1 | 1, l, r, v);
    T[o].v = max(T[o << 1].v, T[o << 1 | 1].v);
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vector<int> tmp;
    for (int(i) = (1); (i) <= (m); (i)++) {
      int id, l, r;
      cin >> id >> l >> r;
      G[id].emplace_back(l, r);
      tmp.emplace_back(l);
      tmp.emplace_back(r);
    }
    sort((tmp).begin(), (tmp).end());
    (tmp).erase(unique((tmp).begin(), (tmp).end()), (tmp).end());
    auto loc = [&](int x) {
      return lower_bound((tmp).begin(), (tmp).end(), x) - tmp.begin() + 1;
    };
    for (int(i) = (1); (i) <= (n); (i)++)
      for (auto &[x, y] : G[i]) {
        x = loc(x);
        y = loc(y);
      }
    build(1, 1, tmp.size());
    for (int(i) = (1); (i) <= (n); (i)++) {
      pair<int, int> cur = {0, -1};
      for (auto [l, r] : G[i]) cur = max(cur, query(1, l, r));
      nxt[i] = cur.second;
      cur.first++, cur.second = i;
      for (auto [l, r] : G[i]) update(1, l, r, cur);
    }
    pair<int, int> mx = query(1, 1, tmp.size());
    cout << n - mx.first << '\n';
    for (int i = mx.second; i != -1; i = nxt[i]) ans[i] = true;
    for (int(i) = (1); (i) <= (n); (i)++)
      if (!ans[i]) cout << i << ' ';
  }
  