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;
  string to_string(const string& s) { return '"' + s + '"'; }
  string to_string(const char* s) { return to_string((string)s); }
  string to_string(bool b) { return (b ? "true" : "false"); }
  template <typename A, typename B>
  string to_string(const pair<A, B>& p);
  template <typename A, typename B, typename C>
  string to_string(tuple<A, B, C> p);
  template <typename T>
  string to_string(const vector<T>& v) {
    string res = "{";
    for (int i = 0; i < static_cast<int>(v.size()); i++) {
      if (i > 0) res += ", ";
      res += to_string(v[i]);
    }
    res += "}";
    return res;
  }
  template <size_t N>
  string to_string(bitset<N> v) {
    string res = "";
    for (size_t i = 0; i < N; i++) {
      res += static_cast<char>('0' + v[i]);
    }
    return res;
  }
  template <typename A>
  string to_string(A v) {
    bool first = true;
    string res = "{";
    for (const auto& x : v) {
      if (!first) res += ", ";
      first = false;
      res += to_string(x);
    }
    res += "}";
    return res;
  }
  template <typename A, typename B>
  string to_string(const pair<A, B>& p) {
    return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
  }
  template <typename A, typename B, typename C>
  string to_string(tuple<A, B, C> p) {
    return "(" + to_string(get<0>(p)) + ", " + to_string(get<1>(p)) + ", " +
           to_string(get<2>(p)) + ")";
  }
  void debug_out() { cerr << '\n'; }
  template <typename Head, typename... Tail>
  void debug_out(Head H, Tail... T) {
    cerr << " " << to_string(H);
    debug_out(T...);
  }
  const int N = 6e5 + 5;
  struct SegmentTree {
    int n;
    struct Node {
      pair<int, int> val, lazy;
    };
    vector<Node> st;
    const Node NIL{{-1, N}, {-1, N}};
    int L(int x) { return (x << 1) + 1; }
    int R(int x) { return (x << 1) + 2; }
    int Mid(int s, int e) { return s + (e - s) / 2; }
    SegmentTree(int n) : n(n) { st.resize(n << 2, NIL); }
    Node merge(const Node& a, const Node& b) {
      return {max(a.val, b.val), NIL.lazy};
    }
    void addLazy(int cur, pair<int, int> lazy) {
      st[cur].lazy = max(st[cur].lazy, lazy);
    }
    void push_down(int cur, int s, int e) {
      st[cur].val = max(st[cur].val, st[cur].lazy);
      if (s != e) {
        addLazy(L(cur), st[cur].lazy);
        addLazy(R(cur), st[cur].lazy);
      }
    }
    void increase(int cur, int s, int e, int i, int j, pair<int, int> newVal) {
      push_down(cur, s, e);
      if (s > j || e < i) return;
      if (s >= i && e <= j) {
        st[cur].lazy = newVal;
        push_down(cur, s, e);
        return;
      }
      int mid = Mid(s, e);
      increase(L(cur), s, mid, i, j, newVal);
      increase(R(cur), mid + 1, e, i, j, newVal);
      st[cur] = merge(st[L(cur)], st[R(cur)]);
    }
    void increase(int l, int r, pair<int, int> newVal) {
      increase(0, 0, n - 1, l, r, newVal);
    }
    Node query(int cur, int s, int e, int i, int j) {
      push_down(cur, s, e);
      if (s > j || e < i) return NIL;
      if (s >= i && e <= j) return st[cur];
      int mid = Mid(s, e);
      return merge(query(L(cur), s, mid, i, j), query(R(cur), mid + 1, e, i, j));
    }
    Node query(int l, int r) { return query(0, 0, n - 1, l, r); }
    void print(int cur, int s, int e) {
      if (s == e) {
        cout << st[cur].val.first << " ";
        return;
      }
      print(L(cur), s, Mid(s, e));
      print(R(cur), Mid(s, e) + 1, e);
    }
    void print() {
      print(0, 0, n - 1);
      cout << '\n';
    }
  };
  int n, m;
  vector<pair<int, int>> row[N];
  pair<int, int> ans[N];
  int exist[N];
  int main() {
    scanf("%d%d", &n, &m);
    map<int, int> id;
    for (int i = 0; i < m; i++) {
      int idx, l, r;
      scanf("%d%d%d", &idx, &l, &r);
      row[--idx].push_back({--l, --r});
      id[l];
      id[r];
    }
    int curId = 0;
    for (auto& x : id) {
      x.second = curId++;
    }
    for (int i = 0; i < n; i++) {
      for (auto& x : row[i]) {
        x.first = id[x.first];
        x.second = id[x.second];
      }
    }
    SegmentTree segTree(curId);
    for (int i = n - 1; i >= 0; i--) {
      ans[i] = {0, n};
      for (int j = 0; j < ((int)(row[i]).size()); j++) {
        auto q = segTree.query(row[i][j].first, row[i][j].second).val;
        q.first++;
        ans[i] = max(ans[i], q);
      }
      for (int j = 0; j < ((int)(row[i]).size()); j++) {
        segTree.increase(row[i][j].first, row[i][j].second, {ans[i].first, i});
      }
    }
    int st = max_element(ans, ans + n) - ans;
    for (int cur = st; cur < n; cur = ans[cur].second) {
      exist[cur] = 1;
    }
    printf("%d\n", n - ans[st].first - 1);
    for (int i = 0; i < n; i++) {
      if (!exist[i]) {
        printf("%d ", i + 1);
      }
    }
    printf("\n");
    return 0;
  }
  