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;
  template <typename T>
  void read(vector<T>& a) {
    for (auto& e : a) cin >> e;
  }
  template <typename T>
  void print(vector<T>& a) {
    for (auto& e : a) cout << e + 1 << ' ';
    cout << '\n';
  }
  vector<pair<int, int> > t, pusher;
  void push(int v, int l, int r) {
    if (v >= t.size()) cout << "push: " << v << ' ' << l << ' ' << r << endl;
    if (pusher[v].second == -1) {
      return;
    }
    t[v] = max(t[v], pusher[v]);
    if (l != r) {
      pusher[v * 2] = max(pusher[v * 2], pusher[v]);
      pusher[v * 2 + 1] = max(pusher[v * 2 + 1], pusher[v]);
    }
    pusher[v] = {0, -1};
  }
  void Set(int v, int l, int r, int ql, int qr, pair<int, int> x) {
    if (v >= t.size()) cout << "set: " << v << ' ' << l << ' ' << r << endl;
    if (qr < l || r < ql) {
      return;
    }
    if (ql <= l && r <= qr) {
      pusher[v] = max(pusher[v], x);
      push(v, l, r);
      return;
    }
    int m = (l + r) / 2;
    Set(v * 2, l, m, ql, qr, x);
    Set(v * 2 + 1, m + 1, r, ql, qr, x);
    t[v] = max(t[v * 2], t[v * 2 + 1]);
  }
  pair<int, int> get(int v, int l, int r, int ql, int qr) {
    if (v >= t.size()) cout << "get: " << v << ' ' << l << ' ' << r << endl;
    push(v, l, r);
    if (qr < l || r < ql) {
      return {0, -1};
    }
    if (ql <= l && r <= qr) {
      return t[v];
    }
    int m = (l + r) / 2;
    return max(get(v * 2, l, m, ql, qr), get(v * 2 + 1, m + 1, r, ql, qr));
  }
  void solve() {
    int n, m;
    cin >> n >> m;
    map<int, vector<pair<int, int> > > lev;
    set<int> c;
    for (int i = 0; i < m; ++i) {
      int ll, l, r;
      cin >> ll >> l >> r;
      --ll;
      --l;
      --r;
      lev[ll].push_back({l, r});
      c.insert(l);
      c.insert(r);
    }
    map<int, int> com;
    auto it = c.begin();
    for (int i = 0; i < c.size(); ++i) {
      com[*(it++)] = i;
    }
    for (auto& el : lev) {
      for (auto& s : el.second) {
        s = {com[s.first], com[s.second]};
      }
    }
    int s = 4 * c.size();
    t.resize(s, {0, -1});
    pusher.resize(s, {0, -1});
    vector<int> prev(n + 1, -1);
    for (auto& el : lev) {
      pair<int, int> mx = {0, -1};
      for (auto& seg : el.second) {
        mx = max(mx, get(1, 0, c.size() - 1, seg.first, seg.second));
      }
      mx.first++;
      prev[el.first] = mx.second;
      mx.second = el.first;
      for (auto& seg : el.second) {
        Set(1, 0, c.size() - 1, seg.first, seg.second, mx);
      }
    }
    int curr = get(1, 0, c.size() - 1, 0, c.size() - 1).second;
    vector<bool> ans(n, 1);
    int cnt = n;
    while (curr != -1) {
      ans[curr] = 0;
      cnt--;
      curr = prev[curr];
    }
    cout << cnt << '\n';
    for (int i = 0; i < n; ++i) {
      if (ans[i]) {
        cout << i + 1 << ' ';
      }
    }
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    solve();
  }
  