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;
  template <class T>
  void _R(T &x) {
    cin >> x;
  }
  void _R(int &x) { scanf("%d", &x); }
  void _R(long long &x) { scanf("%lld", &x); }
  void _R(double &x) { scanf("%lf", &x); }
  void _R(char &x) { scanf("%c", &x); }
  void _R(char *x) { scanf("%s", x); }
  template <class T, class U>
  void _R(pair<T, U> &x) {
    _R(x.first);
    _R(x.second);
  }
  void R() {}
  template <class T, class... U>
  void R(T &head, U &...tail) {
    _R(head);
    R(tail...);
  }
  template <class T>
  void _W(const T &x) {
    cout << x;
  }
  void _W(const unsigned int &x) { printf("%u", x); }
  void _W(const int &x) { printf("%d", x); }
  void _W(const long long &x) { printf("%lld", x); }
  void _W(const double &x) { printf("%.12f", x); }
  void _W(const char &x) { putchar(x); }
  void _W(const char *x) { printf("%s", x); }
  template <class T, class U>
  void _W(const pair<T, U> &x) {
    _W(x.first);
    putchar(' ');
    _W(x.second);
  }
  template <class T>
  void _W(const vector<T> &x) {
    for (auto i = x.begin(); i != x.end(); _W(*i++))
      if (i != x.cbegin()) putchar(' ');
  }
  void W() {}
  template <class T, class... U>
  void W(const T &head, const U &...tail) {
    _W(head);
    putchar(sizeof...(tail) ? ' ' : '\n');
    W(tail...);
  }
  void _E() {}
  template <class T, class... U>
  void _E(const T &head, const U &...tail) {
    _W(head);
    printf(sizeof...(tail) ? " " : " <<-\n");
    _E(tail...);
  }
  void E() {}
  template <class T, class... U>
  void E(const T &head, const U &...tail) {
    printf("〇 ");
    _W(head);
    printf(sizeof...(tail) ? " " : " <<-\n");
    _E(tail...);
  }
  template <class T>
  void A(const T *st, const T *ed) {
    for (auto i = st; i != ed; _W(*i++))
      if (i != st) putchar(' ');
    putchar('\n');
  }
  int n, m;
  pair<int, int> tag[600005 << 2];
  pair<int, int> mx[600005 << 2];
  int id[600005 << 2];
  int hsh[600005], tot;
  int ans[300005];
  int path[300005];
  vector<pair<int, int> > g[300005];
  void build(int i, int l, int r) {
    tag[i] = {0, 0};
    mx[i] = {0, 0};
    if (l == r) {
      mx[i].second = 0;
      return;
    }
    int m = l + r >> 1;
    build(i << 1, l, m);
    build(i << 1 | 1, m + 1, r);
    mx[i] = max(mx[i << 1], mx[i << 1 | 1]);
  }
  void pushdown(int i, int l, int r) {
    if (tag[i].first) {
      tag[i << 1] = tag[i];
      tag[i << 1 | 1] = tag[i];
      mx[i << 1] = tag[i];
      mx[i << 1 | 1] = tag[i];
      tag[i] = {0, 0};
    }
  }
  pair<int, int> query(int i, int l, int r, int L, int R) {
    if (L <= l && r <= R) return mx[i];
    pushdown(i, l, r);
    int m = l + r >> 1;
    pair<int, int> ans = {0, -1};
    if (L <= m) ans = query(i << 1, l, m, L, R);
    if (m < R) ans = max(ans, query(i << 1 | 1, m + 1, r, L, R));
    return ans;
  }
  void update(int i, int l, int r, int L, int R, int v, int id) {
    if (L <= l && r <= R) {
      mx[i] = {v, id};
      tag[i] = {v, id};
      return;
    }
    pushdown(i, l, r);
    int m = l + r >> 1;
    if (L <= m) update(i << 1, l, m, L, R, v, id);
    if (m < R) update(i << 1 | 1, m + 1, r, L, R, v, id);
    mx[i] = max(mx[i << 1], mx[i << 1 | 1]);
  }
  void __Run_Case(int __Case = 0) {
    R(n, m);
    for (int i = 1, op, l, r; i <= m; ++i) {
      R(op, l, r);
      g[op].push_back({l, r});
      hsh[++tot] = l;
      hsh[++tot] = r;
    }
    sort(hsh + 1, hsh + tot + 1);
    tot = unique(hsh + 1, hsh + tot + 1) - hsh - 1;
    int num = 0, id = 0;
    for (int i = 1; i <= n; ++i) {
      pair<int, int> res = {0, -1};
      for (auto &x : g[i]) {
        x.first = lower_bound(hsh + 1, hsh + tot + 1, x.first) - hsh;
        x.second = lower_bound(hsh + 1, hsh + tot + 1, x.second) - hsh;
        res = max(res, query(1, 1, tot, x.first, x.second));
      }
      for (auto x : g[i]) {
        update(1, 1, tot, x.first, x.second, res.first + 1, i);
      }
      ans[i] = res.first + 1;
      path[i] = res.second;
      if (ans[i] > num) {
        num = ans[i];
        id = i;
      }
    }
    num = n - num;
    W(num);
    while (id) {
      ans[id] = -1;
      id = path[id];
    }
    for (int i = 1; i <= n; ++i)
      if (ans[i] != -1) printf("%d%c", i, " \n"[(num--) == 0]);
  }
  void init() {}
  int main() {
    init();
    int __T = 1;
    for (int __Case = 1; __Case <= __T; ++__Case) __Run_Case(__Case);
    return 0;
  }
  