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<int, int> tree[2000000], lazy[2000000];
  int pot(int x) {
    int tmp = 1;
    while (x) {
      tmp *= 2;
      x /= 2;
    }
    return tmp;
  }
  int treeSize;
  void update(int a, int b, pair<int, int> v, int q = 0, int r = treeSize - 1,
              int pt = 1) {
    if (a == q && b == r) {
      if (v > tree[pt]) {
        tree[pt] = v;
      }
      lazy[pt] = max(lazy[pt], v);
      return;
    }
    int mid = (q + r) / 2;
    if (lazy[pt] != make_pair(0, 0)) {
      lazy[pt * 2] = max(lazy[pt * 2], lazy[pt]);
      lazy[pt * 2 + 1] = max(lazy[pt * 2 + 1], lazy[pt]);
      tree[pt * 2] = max(tree[pt * 2], lazy[pt]);
      tree[pt * 2 + 1] = max(tree[pt * 2 + 1], lazy[pt]);
      lazy[pt] = {0, 0};
    }
    if (a <= mid) {
      update(a, min(b, mid), v, q, mid, pt * 2);
    }
    if (b > mid) {
      update(max(a, mid + 1), b, v, mid + 1, r, pt * 2 + 1);
    }
    tree[pt] = max(tree[pt * 2], tree[pt * 2 + 1]);
  }
  pair<int, int> qur(int a, int b, int q = 0, int r = treeSize - 1, int pt = 1) {
    if (a == q && b == r) {
      return tree[pt];
    }
    int mid = (q + r) / 2;
    if (lazy[pt] != make_pair(0, 0)) {
      lazy[pt * 2] = max(lazy[pt * 2], lazy[pt]);
      lazy[pt * 2 + 1] = max(lazy[pt * 2 + 1], lazy[pt]);
      tree[pt * 2] = max(tree[pt * 2], lazy[pt]);
      tree[pt * 2 + 1] = max(tree[pt * 2 + 1], lazy[pt]);
      lazy[pt] = {0, 0};
    }
    pair<int, int> out = {0, 0};
    if (a <= mid) {
      out = max(out, qur(a, min(b, mid), q, mid, pt * 2));
    }
    if (b > mid) {
      out = max(out, qur(max(a, mid + 1), b, mid + 1, r, pt * 2 + 1));
    }
    tree[pt] = max(tree[pt * 2], tree[pt * 2 + 1]);
    return out;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int n, m;
    cin >> n >> m;
    vector<int> conv;
    vector<pair<int, pair<int, int>>> inp;
    for (int i = 0; i < m; i++) {
      int a, b, c;
      cin >> a >> b >> c;
      conv.push_back(b);
      conv.push_back(c);
      inp.push_back({a - 1, {b, c}});
    }
    sort(conv.begin(), conv.end());
    vector<vector<pair<int, int>>> tab(n);
    treeSize = pot(4 + conv.size());
    vector<int> dp(n), poprz(n, -1);
    for (int i = 0; i < inp.size(); i++) {
      int a = lower_bound(conv.begin(), conv.end(), inp[i].second.first) -
              conv.begin();
      int b = lower_bound(conv.begin(), conv.end(), inp[i].second.second) -
              conv.begin();
      tab[inp[i].first].push_back({a, b});
    }
    int wyn = -1;
    int idx = -1;
    for (int i = 0; i < n; i++) {
      pair<int, int> bst = {0, 0};
      for (int j = 0; j < tab[i].size(); j++) {
        bst = max(bst, qur(tab[i][j].first, tab[i][j].second));
      }
      for (int j = 0; j < tab[i].size(); j++) {
        update(tab[i][j].first, tab[i][j].second, {bst.first + 1, i});
      }
      if (bst.first > 0) poprz[i] = bst.second;
      if (bst.first + 1 > wyn) {
        wyn = bst.first + 1;
        idx = i;
      }
    }
    vector<bool> used(n);
    int pt = idx;
    while (pt >= 0) {
      if (used[pt]) break;
      used[pt] = true;
      pt = poprz[pt];
    }
    cout << n - wyn << '\n';
    if (n - wyn > 0) {
      for (int i = 0; i < n; i++) {
        if (!used[i]) cout << i + 1 << " ";
      }
      cout << '\n';
    }
    return 0;
  }
  