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 maxn = 5e5 + 5;
  const int modu = 1e9 + 7;
  const double eps = 1e-9;
  int n, m;
  vector<int> d;
  vector<pair<int, int> > seg[maxn];
  vector<pair<int, int> > maxt, mint, dp;
  vector<bool> tg;
  void pushdown(int o) {
    if (tg[o]) {
      if (maxt[(o << 1)] > maxt[o]) {
        maxt[(o << 1)] = maxt[o];
        mint[(o << 1)] = min(mint[(o << 1)], maxt[o]);
        tg[(o << 1)] = 1;
      }
      if (maxt[((o << 1) | 1)] > maxt[o]) {
        maxt[((o << 1) | 1)] = maxt[o];
        mint[((o << 1) | 1)] = min(mint[((o << 1) | 1)], maxt[o]);
        tg[((o << 1) | 1)] = 1;
      }
      tg[o] = 0;
    }
  }
  void upd(int o, int L, int R, int ql, int qr, pair<int, int> del) {
    if (maxt[o] <= del) return;
    if (ql <= L && R <= qr) {
      maxt[o] = del;
      mint[o] = min(mint[o], del);
      tg[o] = true;
      return;
    }
    pushdown(o);
    int mid = L + R >> 1;
    if (ql <= mid) upd((o << 1), L, mid, ql, qr, del);
    if (qr > mid) upd(((o << 1) | 1), mid + 1, R, ql, qr, del);
    maxt[o] = max(maxt[(o << 1)], maxt[((o << 1) | 1)]);
    mint[o] = min(mint[(o << 1)], mint[((o << 1) | 1)]);
  }
  pair<int, int> qry(int o, int L, int R, int ql, int qr) {
    if (ql <= L && R <= qr) return mint[o];
    pushdown(o);
    pair<int, int> res = make_pair(1e9, 0);
    int mid = L + R >> 1;
    if (ql <= mid) res = min(res, qry((o << 1), L, mid, ql, qr));
    if (qr > mid) res = min(res, qry(((o << 1) | 1), mid + 1, R, ql, qr));
    return res;
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin >> n >> m;
    dp.resize(n + 1, make_pair(0, 0));
    for (int i = 0; i < m; ++i) {
      int x, l, r;
      cin >> x >> l >> r;
      seg[x].push_back(make_pair(l, r));
      d.push_back(l);
      d.push_back(r);
    }
    sort(d.begin(), d.end());
    d.erase(unique(d.begin(), d.end()), d.end());
    maxt.resize(8 * (int)d.size(), make_pair(0, 0));
    mint.resize(8 * (int)d.size(), make_pair(0, 0));
    tg.resize(8 * (int)d.size(), 0);
    int tot = (int)d.size();
    for (int i = 1; i <= n; ++i) {
      pair<int, int> res = make_pair(1e9, 0);
      for (auto qj : seg[i]) {
        int l = lower_bound(d.begin(), d.end(), qj.first) - d.begin() + 1;
        int r = lower_bound(d.begin(), d.end(), qj.second) - d.begin() + 1;
        res = min(res, qry(1, 1, tot, l, r));
      }
      res.first += i - 1;
      dp[i] = res;
      res.first -= i;
      res.second = i;
      for (auto qj : seg[i]) {
        int l = lower_bound(d.begin(), d.end(), qj.first) - d.begin() + 1;
        int r = lower_bound(d.begin(), d.end(), qj.second) - d.begin() + 1;
        upd(1, 1, tot, l, r, res);
      }
    }
    int ans = n;
    for (int i = n - 1; i >= 1; --i)
      if (dp[ans].first + n - ans > dp[i].first + n - i) ans = i;
    cout << dp[ans].first + n - ans << "\n";
    for (int i = n; i > ans; --i) cout << i << " ";
    while (ans) {
      for (int i = ans - 1; i > dp[ans].second; --i) cout << i << " ";
      ans = dp[ans].second;
    }
    return 0;
  }
  