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 maxsize = 600005;
  vector<pair<int, int>> tree(4 * maxsize + 10, {0, -1});
  vector<pair<int, int>> d(4 * maxsize, {-1, -1});
  void push(int p) {
    if (d[p].first != -1) {
      tree[2 * p + 1] =
          (d[p].first > tree[2 * p + 1].first) ? d[p] : tree[2 * p + 1];
      tree[2 * p + 2] =
          (d[p].first > tree[2 * p + 2].first) ? d[p] : tree[2 * p + 2];
      d[2 * p + 1] = d[p];
      d[2 * p + 2] = d[p];
      d[p] = {-1, -1};
    }
  }
  pair<int, int> get_ans(int p, int l, int r, int x, int y) {
    if (l >= y or r <= x) return {0, -1};
    if (l >= x and r <= y) return tree[p];
    int m = (l + r) >> 1;
    push(p);
    auto t1 = get_ans(2 * p + 1, l, m, x, y), t2 = get_ans(2 * p + 2, m, r, x, y);
    return (t1.first > t2.first) ? t1 : t2;
  }
  void rebuild(int p, int l, int r, int x, int y, pair<int, int> v) {
    if (l >= y or r <= x) return;
    if (l >= x and r <= y) {
      tree[p] = (v.first > tree[p].first) ? v : tree[p];
      d[p] = v;
      return;
    }
    int m = (l + r) >> 1;
    push(p);
    rebuild(2 * p + 1, l, m, x, y, v), rebuild(2 * p + 2, m, r, x, y, v);
    tree[p] = (tree[2 * p + 1].first > tree[2 * p + 2].first) ? tree[2 * p + 1]
                                                              : tree[2 * p + 2];
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    vector<pair<int, int>> a[n];
    set<int> s;
    for (int i = 0; i < m; i++) {
      int t, x, y;
      cin >> t >> x >> y;
      t--;
      x--;
      s.insert(x);
      s.insert(y);
      a[t].push_back({x, y});
    }
    map<int, int> mp;
    int tt = 0;
    for (int x : s) mp[x] = tt++;
    for (int i = 0; i < n; i++) {
      for (auto &v : a[i]) v = {mp[v.first], mp[v.second]};
    }
    int ans = 0, en = 0;
    vector<int> predok(n, -1);
    for (int i = 0; i < n; i++) {
      int res = 0;
      for (auto v : a[i]) {
        auto x = get_ans(0, 0, maxsize, v.first, v.second);
        if (x.first + 1 > res) {
          res = x.first + 1;
          predok[i] = x.second;
        }
        if (res > ans) {
          ans = res;
          en = i;
        }
      }
      for (auto v : a[i]) rebuild(0, 0, maxsize, v.first, v.second, {res, i});
    }
    int x = en;
    map<int, bool> q;
    while (x != -1) {
      q[x] = 1;
      x = predok[x];
    }
    cout << n - ans << "\n";
    for (int i = 0; i < n; i++) {
      if (!q[i]) cout << i + 1 << " ";
    }
    return 0;
  }
  