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;
  int n, m, it, l, r, dp[300007], vis[300007], cnt, lst[300007];
  vector<pair<int, int> > vec[300007];
  vector<pair<int, int> > events;
  pair<int, int> t[8 * 300007], par, mod[8 * 300007];
  map<int, int> mp;
  void push(int v) {
    if (mod[v].first > t[2 * v + 1].first) t[2 * v + 1] = mod[v];
    if (mod[v].first > t[2 * v + 2].first) t[2 * v + 2] = mod[v];
    if (mod[v].first > mod[2 * v + 1].first) mod[2 * v + 1] = mod[v];
    if (mod[v].first > mod[2 * v + 2].first) mod[2 * v + 2] = mod[v];
    return;
  }
  void update(int v, int tl, int tr, int L, int R, int val) {
    if (tr < L || R < tl) return;
    if (L <= tl && tr <= R) {
      if (dp[val] > t[v].first) {
        t[v] = {dp[val], val};
      }
      if (dp[val] > mod[v].first) {
        mod[v] = {dp[val], val};
      }
      return;
    }
    push(v);
    int tm = (tl + tr) >> 1;
    update(2 * v + 1, tl, tm, L, R, val);
    update(2 * v + 2, tm + 1, tr, L, R, val);
    if (t[2 * v + 1].first > t[2 * v + 2].first)
      t[v] = t[2 * v + 1];
    else
      t[v] = t[2 * v + 2];
    return;
  }
  pair<int, int> get_mx(int v, int tl, int tr, int L, int R) {
    if (tr < L || R < tl) return {-1, -1};
    if (L <= tl && tr <= R) return t[v];
    push(v);
    int tm = (tl + tr) >> 1;
    pair<int, int> mx1 = get_mx(2 * v + 1, tl, tm, L, R);
    pair<int, int> mx2 = get_mx(2 * v + 2, tm + 1, tr, L, R);
    if (mx1.first == -1) return mx2;
    if (mx2.first == -1) return mx1;
    if (mx1.first > mx2.first)
      return mx1;
    else
      return mx2;
  }
  int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; i++) {
      scanf("%d%d%d", &it, &l, &r);
      vec[it].push_back({l, r});
      events.push_back({l, 1});
      events.push_back({r + 1, 0});
    }
    sort(events.begin(), events.end());
    for (int i = 0; i < 2 * m; i++) {
      l = events[i].first;
      it = events[i].second;
      if (mp[l] == 0) mp[l] = ++cnt;
    }
    for (int i = 1; i <= n; i++) {
      dp[i] = 1;
      for (auto p : vec[i]) {
        l = p.first;
        r = p.second;
        l = mp[l];
        r = mp[r + 1] - 1;
        par = get_mx(0, 1, cnt, l, r);
        if (par.first + 1 > dp[i]) {
          dp[i] = par.first + 1;
          lst[i] = par.second;
        }
      }
      for (auto p : vec[i]) {
        l = p.first;
        r = p.second;
        l = mp[l];
        r = mp[r + 1] - 1;
        update(0, 1, cnt, l, r, i);
      }
    }
    par = {0, 0};
    for (int i = 1; i <= n; i++) {
      if (dp[i] > par.first) {
        par.first = dp[i];
        par.second = i;
      }
    }
    printf("%d\n", n - par.first);
    while (par.second != 0) {
      vis[par.second] = 1;
      par.second = lst[par.second];
    }
    for (int i = 1; i <= n; i++)
      if (vis[i] == 0) printf("%d ", i);
    return 0;
  }
  