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 point {
    int id;
    bool ok;
    int row;
    bool operator<(const point &p) const {
      return (id == p.id) ? ok < p.ok : id < p.id;
    }
  };
  point a[600100];
  vector<int> edge[300100];
  int dp[300100];
  int truoc[300100];
  bool ok[300100];
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
      int x, y, z;
      cin >> x >> y >> z;
      a[i << 1] = {y, 1, x};
      a[i << 1 | 1] = {z + 1, 0, x};
    }
    sort(a, a + 2 * m);
    multiset<int, greater<int> > b;
    multiset<int> c;
    for (int i = 0; i < 2 * m; i++) {
      if (a[i].ok) {
        if (b.upper_bound(a[i].row) != b.end()) {
          edge[*b.upper_bound(a[i].row)].push_back(a[i].row);
        }
        if (c.upper_bound(a[i].row) != c.end()) {
          edge[a[i].row].push_back(*c.upper_bound(a[i].row));
        }
        b.insert(a[i].row);
        c.insert(a[i].row);
      } else {
        b.erase(b.find(a[i].row));
        c.erase(c.find(a[i].row));
      }
    }
    for (int i = 1; i < n; i++) {
      dp[i] = max(dp[i], 1);
      for (auto x : edge[i]) {
        if (dp[x] < dp[i] + 1) {
          dp[x] = dp[i] + 1;
          truoc[x] = i;
        }
      }
    }
    int max1 = -1;
    int id;
    for (int i = 1; i <= n; i++) {
      if (dp[i] > max1) {
        max1 = dp[i];
        id = i;
      }
    }
    cout << n - max1 << '\n';
    int cur = id;
    while (cur != 0) {
      ok[cur] = 1;
      cur = truoc[cur];
    }
    for (int i = 1; i <= n; i++) {
      if (!ok[i]) {
        cout << i << ' ';
      }
    }
    return 0;
  }
  