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;
  using vi = vector<long long>;
  template <typename T>
  std::istream& operator>>(std::istream& input, std::pair<T, T>& data) {
    input >> data.first >> data.second;
    return input;
  }
  template <typename T>
  std::istream& operator>>(std::istream& input, std::vector<T>& data) {
    for (T& first : data) input >> first;
    return input;
  }
  template <typename T>
  std::ostream& operator<<(std::ostream& output, const pair<T, T>& data) {
    output << "(" << data.first << "," << data.second << ")";
    return output;
  }
  template <typename T>
  std::ostream& operator<<(std::ostream& output, const std::vector<T>& data) {
    for (const T& first : data) output << first << " ";
    return output;
  }
  std::ostream& operator<<(std::ostream& output, const __int128& first) {
    __int128 n = first;
    if (n == 0) {
      output << "0";
      return output;
    }
    if (n < 0) {
      n = -n;
      output << "-";
    }
    string s;
    while (n) {
      s += '0' + (n % 10);
      n /= 10;
    }
    reverse(s.begin(), s.end());
    cout << s;
    return output;
  }
  long long div_up(long long a, long long b) {
    return a / b + ((a ^ b) > 0 && a % b);
  }
  long long div_down(long long a, long long b) {
    return a / b - ((a ^ b) < 0 && a % b);
  }
  long long math_mod(long long a, long long b) { return a - b * div_down(a, b); }
  template <class T>
  using V = vector<T>;
  template <class T>
  bool ckmin(T& a, const T& b) {
    return b < a ? a = b, 1 : 0;
  }
  template <class T>
  bool ckmax(T& a, const T& b) {
    return a < b ? a = b, 1 : 0;
  }
  long long gcd(long long a, long long b) {
    while (b) {
      tie(a, b) = make_pair(b, a % b);
    }
    return a;
  }
  long long Bit(long long mask, long long bit) { return (mask >> bit) & 1; }
  const long long N = 6e5 + 7;
  V<pair<long long, long long> > d[N];
  pair<long long, long long> tree[N << 2], best[N << 2];
  void gap(long long v, pair<long long, long long> first) {
    ckmax(tree[v], first);
    ckmax(best[v], first);
  }
  void children(long long v) { best[v] = max(best[v * 2 + 1], best[v * 2 + 2]); }
  void push(long long v) {
    gap(v * 2 + 1, tree[v]);
    gap(v * 2 + 2, tree[v]);
  }
  pair<long long, long long> getmax(long long v, long long l, long long r,
                                    long long ql, long long qr) {
    if (r < ql || qr < l) {
      return make_pair(-N, -N);
    }
    if (ql <= l && r <= qr) {
      return best[v];
    }
    long long m = (l + r) / 2;
    push(v);
    return max(getmax(v * 2 + 1, l, m, ql, qr),
               getmax(v * 2 + 2, m + 1, r, ql, qr));
  }
  pair<long long, long long> getmax(long long l, long long r) {
    return getmax(0, 0, N, l, r);
  }
  void relax(long long v, long long l, long long r, long long ql, long long qr,
             pair<long long, long long> first) {
    if (r < ql || qr < l) {
      return;
    }
    if (ql <= l && r <= qr) {
      gap(v, first);
      return;
    }
    long long m = (l + r) / 2;
    push(v);
    relax(v * 2 + 1, l, m, ql, qr, first);
    relax(v * 2 + 2, m + 1, r, ql, qr, first);
    children(v);
  }
  void relax(long long l, long long r, pair<long long, long long> first) {
    relax(0, 0, N, l, r, first);
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    for (long long i = (0); i < (N << 2); ++i) {
      tree[i] = make_pair(-N, -N);
    }
    long long n, m;
    cin >> n >> m;
    vi c;
    for (long long i = (0); i < (m); ++i) {
      long long row, l, r;
      cin >> row >> l >> r;
      d[row].push_back(make_pair(l, r));
      c.push_back(l);
      c.push_back(r);
    }
    sort(c.begin(), c.end());
    c.resize(unique(c.begin(), c.end()) - c.begin());
    vi dp(n + 1, 1);
    vi par(n + 1, -1);
    pair<long long, long long> ans = make_pair(-1, -1);
    for (long long i = (1); i < (n + 1); ++i) {
      for (auto& seg : d[i]) {
        seg.first = lower_bound(c.begin(), c.end(), seg.first) - c.begin();
        seg.second = lower_bound(c.begin(), c.end(), seg.second) - c.begin();
      }
      for (auto& seg : d[i]) {
        long long l = seg.first, r = seg.second;
        auto t = getmax(l, r);
        t.first++;
        if (ckmax(dp[i], t.first)) {
          par[i] = t.second;
        }
      }
      for (auto& seg : d[i]) {
        long long l = seg.first, r = seg.second;
        relax(l, r, make_pair(dp[i], i));
      }
      ckmax(ans, make_pair(dp[i], i));
    }
    vi used(n + 1);
    long long u = ans.second;
    while (u != -1) {
      used[u] = 1;
      u = par[u];
    }
    cout << n - ans.first << '\n';
    for (long long i = (1); i < (n + 1); ++i) {
      if (!used[i]) {
        cout << i << ' ';
      }
    }
    cout << '\n';
  }
  