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 long long INF = 0x3f3f3f3f;
  const long long MOD = 998244353;
  const long long N = 6e5 + 10;
  struct SEG {
    struct Node {
      long long l, r, mx, idx, cg;
    } t[N << 2];
    void build(long long p, long long l, long long r) {
      t[p].l = l, t[p].r = r;
      t[p].mx = 0, t[p].cg = t[p].idx = -1;
      if (l == r) {
        return;
      }
      long long mid = (l + r) >> 1;
      build((p << 1), l, mid);
      build((p << 1 | 1), mid + 1, r);
    }
    void calc(long long p, long long x, long long id) {
      t[p].mx = t[p].cg = x;
      t[p].idx = id;
    }
    void pushdown(long long p) {
      if (t[p].cg != -1) {
        calc((p << 1), t[p].cg, t[p].idx);
        calc((p << 1 | 1), t[p].cg, t[p].idx);
        t[p].cg = -1;
      }
    }
    void pushup(long long p) {
      t[p].mx = max(t[(p << 1)].mx, t[(p << 1 | 1)].mx);
      t[p].idx = (t[(p << 1)].mx >= t[(p << 1 | 1)].mx) ? t[(p << 1)].idx
                                                        : t[(p << 1 | 1)].idx;
    }
    void update(long long p, long long l, long long r, long long x,
                long long id) {
      if (t[p].l > r || t[p].r < l) return;
      if (t[p].l >= l && t[p].r <= r) {
        calc(p, x, id);
        return;
      }
      pushdown(p);
      update((p << 1), l, r, x, id);
      update((p << 1 | 1), l, r, x, id);
      pushup(p);
    }
    pair<long long, long long> query(long long p, long long l, long long r) {
      if (t[p].l > r || t[p].r < l) return make_pair(0, -1);
      if (t[p].l >= l && t[p].r <= r) {
        return make_pair(t[p].mx, t[p].idx);
      }
      pushdown(p);
      pair<long long, long long> L = query((p << 1), l, r),
                                 R = query((p << 1 | 1), l, r);
      return (L.first >= R.first) ? L : R;
    }
  } seg;
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    long long n, m;
    cin >> n >> m;
    vector<long long> hsh, pre(n, -1);
    vector<vector<pair<long long, long long> > > a(n);
    for (long long i = 0; i < m; i++) {
      long long id, l, r;
      cin >> id >> l >> r;
      a[id - 1].push_back({l, r});
      hsh.push_back(l);
      hsh.push_back(r);
    }
    sort(hsh.begin(), hsh.end());
    hsh.resize(unique(hsh.begin(), hsh.end()) - hsh.begin());
    seg.build(1, 0, hsh.size() - 1);
    for (long long i = 0; i < n; i++) {
      long long mx = 0, prt = -1;
      for (auto &j : a[i]) {
        j.first = lower_bound(hsh.begin(), hsh.end(), j.first) - hsh.begin();
        j.second = lower_bound(hsh.begin(), hsh.end(), j.second) - hsh.begin();
        pair<long long, long long> tmp = seg.query(1, j.first, j.second);
        if (tmp.first > mx) {
          mx = tmp.first;
          prt = tmp.second;
        }
      }
      pre[i] = prt;
      for (auto j : a[i]) {
        seg.update(1, j.first, j.second, mx + 1, i);
      }
    }
    cout << n - seg.t[1].mx << '\n';
    vector<long long> vis(n);
    for (long long i = seg.t[1].idx; i >= 0; i = pre[i]) {
      vis[i] = 1;
    }
    for (long long i = 0; i < n; i++) {
      if (!vis[i]) {
        cout << i + 1 << ' ';
      }
    }
    cout << '\n';
    return 0;
  }
  