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;
  #pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-poller", "inline", \
                       "03")
  const long double PI = acos(-1);
  const long double eps = 1e-6;
  const long double pi = acos((long double)-1);
  const int MAX = 1e6 + 5;
  const int mod = 1e9 + 7;
  pair<int, int> st[4 * MAX], lz[4 * MAX];
  pair<int, int> ask(pair<int, int> &a, pair<int, int> &b) {
    return a > b ? a : b;
  }
  void push(int x, pair<int, int> v) {
    if (x < 4 * MAX) {
      st[x] = ask(st[x], v);
      lz[x] = ask(lz[x], v);
    }
  }
  void update(int x, int L, int R, int lo, int hi, pair<int, int> v) {
    if (lo > R || hi < L) return;
    if (lo <= L && R <= hi) {
      push(x, v);
      return;
    }
    if (lz[x].first) {
      push(2 * x, lz[x]);
      push(2 * x + 1, lz[x]);
      lz[x] = pair<int, int>(0, 0);
    }
    int mid = (L + R) / 2;
    update(2 * x, L, mid, lo, hi, v);
    update(2 * x + 1, mid + 1, R, lo, hi, v);
    st[x] = ask(st[2 * x], st[2 * x + 1]);
  }
  pair<int, int> query(int x, int L, int R, int lo, int hi) {
    if (lo > R || hi < L) return pair<int, int>(0, 0);
    if (lo <= L && R <= hi) {
      return st[x];
    }
    if (lz[x].first) {
      push(2 * x, lz[x]);
      push(2 * x + 1, lz[x]);
      lz[x] = pair<int, int>(0, 0);
    }
    int mid = (L + R) / 2;
    pair<int, int> v1 = query(2 * x, L, mid, lo, hi);
    pair<int, int> v2 = query(2 * x + 1, mid + 1, R, lo, hi);
    return ask(v1, v2);
  }
  int t, n, q, a[MAX], b[MAX], c[MAX];
  pair<int, int> sol[MAX];
  vector<pair<int, int> > pos[MAX];
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> q;
    vector<int> X;
    for (int i = 1; i <= q; i++) {
      cin >> a[i] >> b[i] >> c[i];
      X.push_back(b[i]);
      X.push_back(c[i]);
    }
    sort((X).begin(), (X).end());
    X.erase(unique((X).begin(), (X).end()), X.end());
    for (int i = 1; i <= q; i++) {
      int x = upper_bound((X).begin(), (X).end(), b[i]) - X.begin();
      int y = upper_bound((X).begin(), (X).end(), c[i]) - X.begin();
      pos[a[i]].push_back(pair<int, int>(x, y));
    }
    int L = X.size();
    for (int i = 1; i <= n; i++) {
      pair<int, int> &best = sol[i];
      for (auto [x, y] : pos[i]) {
        pair<int, int> aux = query(1, 1, L, x, y);
        if (aux > best) best = aux;
      }
      for (auto [x, y] : pos[i])
        update(1, 1, L, x, y, pair<int, int>(best.first + 1, i));
    }
    int P = 1;
    for (int i = 2; i <= n; i++)
      if (sol[i] > sol[P]) P = i;
    vector<int> mark(n + 1);
    int cant = n;
    while (P) {
      mark[P] = 1;
      P = sol[P].second;
      cant--;
    }
    cout << cant << '\n';
    for (int i = 1; i <= n; i++) {
      if (!mark[i]) {
        cout << i << " \n"[--cant == 0];
      }
    }
  }
  