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 N = 600005;
  vector<array<int, 2>> tree(4 * N, {0, 0});
  vector<array<int, 2>> lazy(4 * N, {0, 0});
  void apply(int idx, array<int, 2> a) {
    tree[idx] = max(tree[idx], a);
    lazy[idx] = max(lazy[idx], a);
  }
  void push(int idx) {
    for (int i = 0; i < 2; i++) apply(2 * idx + i, lazy[idx]);
  }
  void update(int idx, int l, int r, int ql, int qr, array<int, 2> a) {
    if (ql > qr) return;
    if (l == ql && r == qr)
      apply(idx, a);
    else {
      int m = (l + r) / 2;
      push(idx);
      update(2 * idx, l, m, ql, min(qr, m), a);
      update(2 * idx + 1, m + 1, r, max(ql, m + 1), qr, a);
      tree[idx] = max(tree[2 * idx], tree[2 * idx + 1]);
    }
  }
  array<int, 2> query(int idx, int l, int r, int ql, int qr) {
    if (ql > qr) return {0, 0};
    if (l == ql && r == qr) return tree[idx];
    int m = (l + r) / 2;
    push(idx);
    return max(query(2 * idx, l, m, ql, min(qr, m)),
               query(2 * idx + 1, m + 1, r, max(ql, m + 1), qr));
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    vector<array<int, 2>> v[n + 1];
    vector<array<int, 4>> h;
    for (int i = 0; i < m; i++) {
      int x, l, r;
      cin >> x >> l >> r;
      h.push_back({l, x, (int)v[x].size(), 0});
      h.push_back({r, x, (int)v[x].size(), 1});
      v[x].push_back({l, r});
    }
    sort(h.begin(), h.end());
    int id = 1;
    for (int i = 0; i < 2 * m; i++) {
      if (i > 0 && h[i][0] != h[i - 1][0]) id++;
      v[h[i][1]][h[i][2]][h[i][3]] = id;
    }
    vector<array<int, 2>> best(n + 1, {0, 0});
    array<int, 2> opt = {0, 0};
    for (int i = 1; i <= n; i++) {
      for (auto [l, r] : v[i]) best[i] = max(best[i], query(1, 1, id, l, r));
      best[i][0]++;
      for (auto [l, r] : v[i]) update(1, 1, id, l, r, {best[i][0], i});
      opt = max(opt, {best[i][0], i});
    }
    vector<bool> vis(n + 1, 0);
    int x = opt[1];
    while (x > 0) {
      vis[x] = 1;
      x = best[x][1];
    }
    cout << n - opt[0] << "\n";
    for (int i = 1; i <= n; i++)
      if (vis[i] == 0) cout << i << " ";
    cout << "\n";
    return 0;
  }
  