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>
  #pragma GCC optimize("Ofast")
  using namespace std;
  using ll = long long;
  using ull = unsigned long long;
  using ld = long double;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  ll getrnd(ll l, ll r) { return uniform_int_distribution<ll>(l, r)(rng); }
  template <typename T1, typename T2>
  inline bool relax(T1& a, const T2& b) {
    return a > b ? a = b, true : false;
  }
  template <typename T1, typename T2>
  inline bool strain(T1& a, const T2& b) {
    return a < b ? a = b, true : false;
  }
  const int N = 6e5 + 5;
  pair<int, int> tree[4 * N], lazy[4 * N];
  pair<int, int> merge(pair<int, int> a, pair<int, int> b) {
    if (a.first > b.first) return a;
    return b;
  }
  void upd(int v, pair<int, int> x) {
    tree[v] = merge(tree[v], x);
    lazy[v] = merge(lazy[v], x);
  }
  void push(int v) {
    if (lazy[v].first >= 0) {
      upd(2 * v, lazy[v]);
      upd(2 * v + 1, lazy[v]);
      lazy[v].first = -1;
    }
  }
  void build(int v, int tl, int tr) {
    if (tl == tr)
      tree[v] = {-1, tl}, lazy[v] = {-1, -1};
    else {
      int tm = (tl + tr) / 2;
      build(2 * v, tl, tm);
      build(2 * v + 1, tm + 1, tr);
      tree[v] = merge(tree[2 * v], tree[2 * v + 1]);
    }
  }
  void update(int v, int tl, int tr, int l, int r, pair<int, int> x) {
    if (l > r) return;
    if (tl == l && tr == r)
      upd(v, x);
    else {
      int tm = (tl + tr) / 2;
      push(v);
      update(2 * v, tl, tm, l, min(r, tm), x);
      update(2 * v + 1, tm + 1, tr, max(l, tm + 1), r, x);
      tree[v] = merge(tree[2 * v], tree[2 * v + 1]);
    }
  }
  pair<int, int> get(int v, int tl, int tr, int l, int r) {
    if (l > r) return {-1, -1};
    if (tl == l && tr == r)
      return tree[v];
    else {
      int tm = (tl + tr) / 2;
      push(v);
      return merge(get(2 * v, tl, tm, l, min(r, tm)),
                   get(2 * v + 1, tm + 1, tr, max(l, tm + 1), r));
    }
  }
  void solve() {
    int n, m;
    cin >> n >> m;
    vector<int> ids;
    vector<vector<pair<int, int>>> segs(n);
    for (int i = 0; i < m; ++i) {
      int ind, l, r;
      cin >> ind >> l >> r;
      --ind;
      ids.emplace_back(l);
      ids.emplace_back(r);
      segs[ind].emplace_back(l, r);
    }
    sort(ids.begin(), ids.end());
    ids.resize(unique(ids.begin(), ids.end()) - ids.begin());
    auto gid = [&](int x) {
      return lower_bound(ids.begin(), ids.end(), x) - ids.begin();
    };
    int sz = ids.size();
    build(1, 0, sz);
    vector<int> dp(n, 1), from(n, -1);
    for (int i = 0; i < n; ++i) {
      for (auto& [l, r] : segs[i]) {
        l = gid(l);
        r = gid(r);
        auto [val, ind] = get(1, 0, sz, l, r);
        if (strain(dp[i], val + 1)) from[i] = ind;
      }
      for (auto [l, r] : segs[i]) {
        update(1, 0, sz, l, r, {dp[i], i});
      }
    }
    int st = max_element(dp.begin(), dp.end()) - dp.begin();
    cout << n - dp[st] << '\n';
    vector<int> was(n);
    while (st != -1) {
      was[st] = 1;
      st = from[st];
    }
    for (int i = 0; i < n; ++i)
      if (!was[i]) cout << i + 1 << ' ';
    cout << '\n';
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int t = 1;
    while (t--) solve();
    return 0;
  }
  