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 = 1e6 + 10;
  int cnt = 0;
  bool f[Maxn];
  struct segmentree {
    int Max[4 * Maxn];
    int lazy[4 * Maxn];
    int Id[4 * Maxn];
    segmentree() {
      memset(Max, 0, sizeof Max);
      memset(lazy, 0, sizeof lazy);
    }
    void Down(int x) {
      if (lazy[x] == 0) return;
      for (int j = 2 * x; j <= 2 * x + 1; j++)
        Max[j] = lazy[x], lazy[j] = lazy[x], Id[j] = Id[x];
      lazy[x] = 0;
    }
    void update(int l, int r, int x, int ql, int qr, int val, int id) {
      if (r < ql || l > qr) return;
      if (l >= ql && r <= qr) {
        Max[x] = val;
        lazy[x] = val;
        Id[x] = id;
        return;
      }
      int mid = l + r >> 1;
      Down(x);
      update(l, mid, x * 2, ql, qr, val, id);
      update(mid + 1, r, x * 2 + 1, ql, qr, val, id);
      Max[x] = max(Max[x * 2], Max[x * 2 + 1]);
      if (Max[x] == Max[2 * x])
        Id[x] = Id[2 * x];
      else
        Id[x] = Id[2 * x + 1];
    }
    pair<int, int> Get(int l, int r, int x, int ql, int qr) {
      if (l > qr || r < ql) return {0, 0};
      if (l >= ql && r <= qr) return {Max[x], Id[x]};
      Down(x);
      int mid = l + r >> 1;
      return max(Get(l, mid, x * 2, ql, qr), Get(mid + 1, r, x * 2 + 1, ql, qr));
    }
  } Tree;
  vector<pair<int, int> > a[Maxn];
  int trace[Maxn];
  int n, m;
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    map<int, int> d;
    for (int i = 1; i <= m; i++) {
      int k, l, r;
      cin >> k >> l >> r;
      a[k].push_back({l, r});
      d[l] = d[r] = 1;
    }
    for (auto &x : d) {
      x.second = ++cnt;
    }
    pair<int, int> ans = {0, 0};
    for (int i = 1; i <= n; i++) {
      pair<int, int> Max = {0, 0};
      for (auto seg : a[i]) {
        int l = d[seg.first];
        int r = d[seg.second];
        pair<int, int> x = Tree.Get(1, cnt, 1, l, r);
        x.first++;
        Max = max(Max, x);
      }
      ans = max(ans, {Max.first, i});
      trace[i] = Max.second;
      for (auto seg : a[i]) {
        int l = d[seg.first];
        int r = d[seg.second];
        Tree.update(1, cnt, 1, l, r, Max.first, i);
      }
    }
    int s = ans.second;
    while (s) {
      f[s] = true;
      s = trace[s];
    }
    cout << n - ans.first << "\n";
    for (int i = 1; i <= n; i++) {
      if (!f[i]) cout << i << " ";
    }
    return 0;
  }
  