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;
  struct segment {
    int l, u;
  } tmp;
  pair<int, int> seg[4 * int(6e5 + 10)], lazy[4 * int(6e5 + 10)], x;
  int p[int(6e5 + 10)], n, m, k, dp;
  pair<int, int> ans = {-1, -1};
  vector<segment> g[int(6e5 + 10)];
  set<int> s;
  map<int, int> mp;
  bool vis[int(6e5 + 10)];
  pair<int, int> fmax(pair<int, int> a, pair<int, int> b) {
    if (a.first < b.first) return b;
    return a;
  }
  void build(int l, int h, int pos) {
    if (l == h) {
      seg[pos] = lazy[pos] = {0, 0};
      return;
    }
    int mid = (l + h) / 2;
    build(l, mid, 2 * pos);
    build(mid + 1, h, 2 * pos + 1);
    seg[pos] = lazy[pos] = {0, 0};
  }
  void update(int l, int h, int x, int y, pair<int, int> z, int pos) {
    if (lazy[pos].first) {
      seg[pos] = fmax(seg[pos], lazy[pos]);
      if (l != h) {
        lazy[2 * pos] = fmax(lazy[pos], lazy[2 * pos]);
        lazy[2 * pos + 1] = fmax(lazy[pos], lazy[2 * pos + 1]);
      }
      lazy[pos] = {0, 0};
    }
    if (h < x || l > y) return;
    if (x <= l && h <= y) {
      seg[pos] = fmax(seg[pos], z);
      if (l != h) {
        lazy[2 * pos] = fmax(lazy[2 * pos], z);
        lazy[2 * pos + 1] = fmax(lazy[2 * pos], z);
      }
      return;
    }
    int mid = (l + h) / 2;
    update(l, mid, x, y, z, 2 * pos);
    update(mid + 1, h, x, y, z, 2 * pos + 1);
    seg[pos] = fmax(seg[2 * pos], seg[2 * pos + 1]);
  }
  pair<int, int> query(int l, int h, int x, int y, int pos) {
    if (lazy[pos].first) {
      seg[pos] = fmax(seg[pos], lazy[pos]);
      if (l != h) {
        lazy[2 * pos] = fmax(lazy[pos], lazy[2 * pos]);
        lazy[2 * pos + 1] = fmax(lazy[pos], lazy[2 * pos + 1]);
      }
      lazy[pos] = {0, 0};
    }
    if (h < x || l > y) return {-1, -1};
    if (x <= l && h <= y) {
      return seg[pos];
    }
    int mid = (l + h) / 2;
    return fmax(query(l, mid, x, y, 2 * pos),
                query(mid + 1, h, x, y, 2 * pos + 1));
  }
  void fprint(int t) {
    if (p[t] != t) {
      fprint(p[t]);
    }
    vis[t] = true;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    cout.precision(10);
    cin >> n >> m;
    for (int i = 1; i <= m; i++) {
      cin >> k >> tmp.l >> tmp.u;
      s.insert(tmp.l);
      s.insert(tmp.u);
      g[k].push_back(tmp);
    }
    k = 1;
    for (auto x : s) {
      if (mp.find(x) == mp.end()) {
        mp[x] = k++;
      }
    }
    build(1, k, 1);
    for (int i = 1; i <= n; i++) {
      p[i] = i;
      dp = 1;
      for (auto s : g[i]) {
        x = query(1, k, mp[s.l], mp[s.u], 1);
        if (dp < 1 + x.first) {
          p[i] = x.second;
          dp = 1 + x.first;
        }
      }
      for (auto s : g[i]) {
        update(1, k, mp[s.l], mp[s.u], make_pair(dp, i), 1);
      }
      ans = fmax(ans, make_pair(dp, i));
    }
    cout << n - ans.first << endl;
    fprint(ans.second);
    for (int i = 1; i <= n; i++) {
      if (!vis[i]) cout << i << ' ';
    }
    cout << endl;
    return 0;
  }
  