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 segment_tree {
    int n;
    vector<pair<int, int>> st, lazy;
    segment_tree(int n) : n(n), st(2 * n, {0, -1}), lazy(2 * n) {}
    inline int id(int b, int e) { return (b + e - 1) | (b != e - 1); }
    void prop(int l, int r) {
      int cur = id(l, r);
      st[cur] = max(st[cur], lazy[cur]);
      if (l + 1 != r) {
        int mid = (l + r + 1) >> 1;
        lazy[id(l, mid)] = max(lazy[cur], lazy[id(l, mid)]);
        lazy[id(mid, r)] = max(lazy[id(l, mid)], lazy[cur]);
      }
      lazy[cur] = {0, -1};
    }
    void upd(int li, int ri, pair<int, int> v) { upd(0, n, li, ri, v); }
    void upd(int l, int r, int li, int ri, pair<int, int> v) {
      if (li == ri) return;
      int cur = id(l, r);
      if (lazy[cur].first) prop(l, r);
      if (l >= ri || r <= li) return;
      if (li <= l && r <= ri) {
        lazy[cur] = v;
        prop(l, r);
        return;
      }
      int mid = (l + r + 1) >> 1;
      upd(l, mid, li, ri, v);
      upd(mid, r, li, ri, v);
      st[cur] = max(st[id(l, mid)], st[id(mid, r)]);
    }
    pair<int, int> que(int li, int ri) { return que(0, n, li, ri); }
    pair<int, int> que(int l, int r, int li, int ri) {
      int cur = id(l, r);
      if (lazy[cur].first) prop(l, r);
      if (l >= ri || r <= li) return {0, -1};
      if (li <= l && r <= ri) return st[cur];
      int mid = (l + r + 1) >> 1;
      return max(que(l, mid, li, ri), que(mid, r, li, ri));
    }
  };
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    vector<int> w;
    vector<vector<pair<int, int>>> e(n);
    for (int i = 0; i < m; i++) {
      int t, l, r;
      cin >> t >> l >> r;
      e[t - 1].push_back({l, r + 1});
      w.push_back(l);
      w.push_back(r + 1);
    }
    sort(w.begin(), w.end());
    w.resize(unique(w.begin(), w.end()) - w.begin());
    int sz = w.size();
    map<int, int> mp;
    for (int i = 0; i < sz; i++) mp[w[i]] = i;
    for (int i = 0; i < n; i++)
      for (auto &j : e[i]) j.first = mp[j.first], j.second = mp[j.second];
    segment_tree st(sz);
    vector<int> p(n, -1);
    for (int i = 0; i < n; i++) {
      pair<int, int> mx = {0, -1};
      for (auto j : e[i]) {
        pair<int, int> t = st.que(j.first, j.second);
        mx = max(mx, t);
      }
      if (mx.first == 0) mx.second = -1;
      p[i] = mx.second;
      for (auto j : e[i]) st.upd(j.first, j.second, {mx.first + 1, i});
    }
    int y = st.que(0, sz).second;
    vector<bool> in(n);
    while (y != -1) {
      in[y] = true;
      y = p[y];
    }
    vector<int> ans;
    for (int i = 0; i < n; i++)
      if (!in[i]) ans.push_back(i);
    cout << ans.size() << "\n";
    for (auto &i : ans) cout << i + 1 << " \n"[&i == &ans.back()];
    return 0;
  }
  