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 off = 1 << 20, maxn = 3e5;
  int n, m, k, l, r;
  vector<int> v;
  vector<pair<int, int> > inte[maxn];
  pair<int, int> tour[2 * off];
  pair<int, int> prop[2 * off];
  pair<int, int> dp[maxn];
  int bio[maxn];
  void f(int x, int v, int ind) {
    if (v > tour[x].first) {
      tour[x] = {v, ind};
    }
    if (v > prop[x].first) {
      prop[x] = {v, ind};
    }
  }
  void push(int x) {
    if (x >= off || prop[x].first == 0) return;
    int v = prop[x].first;
    int ind = prop[x].second;
    prop[x].first = 0;
    f(2 * x, v, ind);
    f(2 * x + 1, v, ind);
  }
  void update(int x, int lo, int hi, int l, int r, int v, int ind) {
    push(x);
    if (hi <= l || r <= lo) return;
    if (l <= lo && hi <= r) {
      f(x, v, ind);
      return;
    }
    int mid = (lo + hi) / 2;
    update(x * 2, lo, mid, l, r, v, ind);
    update(x * 2 + 1, mid, hi, l, r, v, ind);
    pair<int, int> t1 = tour[x * 2];
    pair<int, int> t2 = tour[x * 2 + 1];
    if (t1.first > t2.first)
      tour[x] = t1;
    else
      tour[x] = t2;
  }
  pair<int, int> query(int x, int lo, int hi, int l, int r) {
    push(x);
    if (hi <= l || r <= lo) return {0, -1};
    if (l <= lo && hi <= r) return tour[x];
    int mid = (lo + hi) / 2;
    pair<int, int> q1 = query(x * 2, lo, mid, l, r);
    pair<int, int> q2 = query(x * 2 + 1, mid, hi, l, r);
    if (q1.first > q2.first) return q1;
    return q2;
  }
  void ispis() {
    for (int i = 1; i < 2 * off; i++) {
      if ((i & (i - 1)) == 0) cout << endl;
      cout << "[" << tour[i].first << ' ' << prop[i].first << ' '
           << tour[i].second << "] ";
    }
    cout << endl;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
      cin >> k >> l >> r;
      k--;
      inte[k].push_back({l, r});
      v.push_back(l);
      v.push_back(r);
    }
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end());
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < inte[i].size(); j++) {
        inte[i][j].first =
            lower_bound(v.begin(), v.end(), inte[i][j].first) - v.begin();
        inte[i][j].second =
            lower_bound(v.begin(), v.end(), inte[i][j].second) - v.begin();
      }
    }
    for (int i = 1; i < 2 * off; i++) tour[i].second = -1;
    for (int i = 0; i < n; i++) {
      dp[i].second = -1;
      for (int j = 0; j < inte[i].size(); j++) {
        pair<int, int> q =
            query(1, 0, off, inte[i][j].first, inte[i][j].second + 1);
        if (q.first > dp[i].first) dp[i] = q;
      }
      dp[i].first++;
      for (int j = 0; j < inte[i].size(); j++) {
        update(1, 0, off, inte[i][j].first, inte[i][j].second + 1, dp[i].first,
               i);
      }
    }
    int x = n;
    for (int i = 0; i < n; i++) {
      if (dp[i].first > dp[x].first) x = i;
    }
    cout << n - dp[x].first << "\n";
    while (x != -1) {
      bio[x] = 1;
      x = dp[x].second;
    }
    for (int i = 0; i < n; i++) {
      if (!bio[i]) cout << i + 1 << ' ';
    }
    return 0;
  }
  