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 long long INF = (long long)1 << 62;
  const long long MOD = 1e9 + 7;
  const int iINF = 1 << 30;
  const double PI = 3.14159265359;
  struct intv {
    int i, tmx, tpe;
  };
  bool cmp(intv a, intv b) {
    if (a.tmx == b.tmx) {
      if (a.tpe == b.tpe) return a.i < b.i;
      return a.tpe < b.tpe;
    }
    return a.tmx < b.tmx;
  }
  void solve() {
    int n, m;
    cin >> n >> m;
    vector<intv> a;
    for (int i = 0; i < m; i++) {
      int x, l, r;
      cin >> x >> l >> r;
      a.push_back(intv{x - 1, l, 0});
      a.push_back(intv{x - 1, r, 1});
    }
    sort(a.begin(), a.end(), cmp);
    vector<vector<int> > adj(n);
    map<int, int> active;
    for (int i = 0; i < (int)a.size(); i++) {
      if (a[i].tpe == 0) {
        if (active.find(a[i].i) == active.end()) active[a[i].i] = 0;
        active[a[i].i]++;
        auto it = active.lower_bound(a[i].i);
        it++;
        if (it != active.end()) adj[a[i].i].push_back(it->first);
        it--;
        if (it != active.begin()) {
          it--;
          adj[it->first].push_back(a[i].i);
        }
      } else {
        if (active[a[i].i] == 1) {
          auto it = active.lower_bound(a[i].i);
          it++;
          int u = -1, v = -1;
          if (it != active.end()) v = it->first;
          it--;
          if (it != active.begin()) {
            it--;
            u = it->first;
          }
          if (u != -1 && v != -1) adj[u].push_back(v);
        }
        active[a[i].i]--;
        if (active[a[i].i] == 0) active.erase(active.find(a[i].i));
      }
    }
    vector<int> dp(n, 1);
    int strt = 0;
    for (int u = n - 1; u >= 0; u--) {
      for (int v : adj[u]) dp[u] = max(dp[u], dp[v] + 1);
      if (dp[u] > dp[strt]) strt = u;
    }
    set<int> taken;
    int curr = strt;
    taken.insert(curr);
    while ((int)taken.size() != dp[strt]) {
      int maxi = adj[curr][0];
      for (int v : adj[curr])
        if (dp[v] > dp[maxi]) maxi = v;
      curr = maxi;
      taken.insert(curr);
    }
    cout << n - taken.size() << "\n";
    for (int i = 0; i < n; i++) {
      if (taken.find(i) == taken.end()) cout << i + 1 << " ";
    }
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  