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;
  pair<long long, long long> t[8 * 300009], lazy[8 * 300009];
  long long x[300009], l[300009], r[300009], dp[300009], par[300009];
  vector<long long> ind[300009];
  void push(long long v) {
    if (lazy[v].first == 0) return;
    if (lazy[v].first >= t[v + v].first) t[v + v] = lazy[v + v] = lazy[v];
    if (lazy[v].first >= t[v + v + 1].first)
      t[v + v + 1] = lazy[v + v + 1] = lazy[v];
    lazy[v * 2 + 1] = lazy[v];
    lazy[v] = {0, 0};
  }
  void update(long long v, long long tl, long long tr, long long l, long long r,
              pair<long long, long long> addend) {
    if (l > r) return;
    if (l == tl && tr == r) {
      if (addend.first >= t[v].first) {
        t[v] = lazy[v] = addend;
      }
    } else {
      push(v);
      long long tm = (tl + tr) / 2;
      update(v * 2, tl, tm, l, min(r, tm), addend);
      update(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, addend);
      if (t[v + v].first > t[v + v + 1].first)
        t[v] = t[v * 2];
      else
        t[v] = t[v + v + 1];
    }
  }
  pair<long long, long long> query(long long v, long long tl, long long tr,
                                   long long l, long long r) {
    if (l > r) return {0, 0};
    if (l <= tl && tr <= r) return t[v];
    push(v);
    long long tm = (tl + tr) / 2;
    pair<long long, long long> a = query(v * 2, tl, tm, l, min(r, tm));
    pair<long long, long long> b =
        query(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
    if (a.first >= b.first) return a;
    return b;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    vector<long long> v;
    long long n, m;
    cin >> n >> m;
    long long i = 0;
    while (i < m) {
      cin >> x[i] >> l[i] >> r[i];
      v.push_back(l[i]);
      v.push_back(r[i]);
      ind[x[i]].push_back(i);
      i++;
    }
    sort((v).begin(), (v).end());
    v.resize(unique((v).begin(), (v).end()) - v.begin());
    i = 0;
    long long mx = 0;
    while (i < m) {
      l[i] = 1 + lower_bound(v.begin(), v.end(), l[i]) - v.begin();
      r[i] = 1 + lower_bound(v.begin(), v.end(), r[i]) - v.begin();
      mx = max({mx, l[i], r[i]});
      i++;
    }
    long long op = 0;
    pair<long long, long long> temp;
    i = 1;
    while (i <= n) {
      dp[i] = 1;
      for (auto s : ind[i]) {
        temp = query(1, 1, mx, l[s], r[s]);
        if (1 + temp.first >= dp[i]) {
          dp[i] = 1 + temp.first;
          par[i] = temp.second;
        }
      }
      for (auto s : ind[i]) {
        update(1, 1, mx, l[s], r[s], {dp[i], i});
      }
      if (dp[i] > dp[op]) op = i;
      i++;
    }
    cout << n - dp[op] << "\n";
    set<long long> s;
    while (op) {
      s.insert(op);
      op = par[op];
    }
    i = 1;
    while (i <= n) {
      if (s.find(i) == s.end()) cout << i << " ";
      i++;
    }
  }
  