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;
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vector<int> cover(n);
    vector<tuple<int, int, int>> endpoints;
    set<int> s{-1, n};
    for (int i = 0; i < m; i++) {
      int x, l, r;
      cin >> x >> l >> r;
      x--;
      l--;
      endpoints.emplace_back(l, x, -1);
      endpoints.emplace_back(r, x, 1);
    }
    sort(endpoints.begin(), endpoints.end());
    vector<pair<int, int>> edges;
    vector<int> add, del;
    for (int l = 0, r; l < 2 * m; l = r) {
      r = l;
      while (r < 2 * m && get<0>(endpoints[l]) == get<0>(endpoints[r])) r++;
      add.clear();
      del.clear();
      for (int i = l; i < r; i++) {
        auto [v, x, t] = endpoints[i];
        if (t == -1) {
          if (cover[x] == 0) add.push_back(x);
          cover[x]++;
        } else {
          cover[x]--;
          if (cover[x] == 0) del.push_back(x);
        }
      }
      for (int x : del) s.erase(x);
      for (int x : add) {
        auto it = s.insert(x).first;
        edges.emplace_back(*prev(it), x);
        edges.emplace_back(x, *next(it));
      }
    }
    sort(edges.begin(), edges.end());
    vector<int> dp(n + 2), g(n + 2);
    for (auto [u, v] : edges) {
      if (dp[u + 1] + 1 > dp[v + 1]) {
        dp[v + 1] = dp[u + 1] + 1;
        g[v + 1] = u + 1;
      }
    }
    int res = n + 1 - dp[n + 1];
    cout << res << "\n";
    vector<int> ans;
    for (int i = n + 1; i > 0; i = g[i]) {
      for (int j = g[i] + 1; j < i; j++) {
        ans.push_back(j);
      }
    }
    for (int i = 0; i < res; i++) cout << ans[i] << " \n"[i == res - 1];
    return 0;
  }
  