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;
  void FAST_IO(string filein = "", string fileout = "") {
    if (fopen(filein.c_str(), "r")) {
      freopen(filein.c_str(), "r", stdin);
      freopen(fileout.c_str(), "w", stdout);
    }
    cin.tie(0), cout.tie(0)->sync_with_stdio(0);
  }
  void Why_Are_You_So_Orz();
  signed main() {
    FAST_IO("mike4235orz.inp", "mike4235orz.out");
    int testcases = 1;
    for (int test = 1; test <= testcases; test++) Why_Are_You_So_Orz();
    return 0;
  }
  const int N = 3e5 + 5;
  int n, m, dp[N], tr[N], cnt;
  vector<pair<int, int>> q[N];
  pair<int, int> st[N << 3], lz[N << 3];
  void apply(int id, pair<int, int> v) {
    lz[id] = max(lz[id], v);
    st[id] = max(st[id], v);
  }
  void push(int id) { apply(id << 1, lz[id]), apply(id << 1 | 1, lz[id]); }
  void update(int l, int r, pair<int, int> v, int id = 1, int tl = 1,
              int tr = cnt) {
    if (l > tr || tl > r) return;
    if (l <= tl && tr <= r) return apply(id, v);
    int tm = tl + tr >> 1;
    push(id);
    update(l, r, v, id << 1, tl, tm), update(l, r, v, id << 1 | 1, tm + 1, tr);
    st[id] = max(st[id << 1], st[id << 1 | 1]);
  }
  pair<int, int> query(int l, int r, int id = 1, int tl = 1, int tr = cnt) {
    if (l > tr || tl > r) return {0, 0};
    if (l <= tl && tr <= r) return st[id];
    int tm = tl + tr >> 1;
    push(id);
    return max(query(l, r, id << 1, tl, tm),
               query(l, r, id << 1 | 1, tm + 1, tr));
  }
  int chk[N];
  void Why_Are_You_So_Orz() {
    cin >> n >> m;
    map<int, int> mp;
    for (int i = 1, v, l, r; i <= m; i++) {
      cin >> v >> l >> r, mp[l], mp[r], q[v].emplace_back(l, r);
    }
    for (auto &vp : mp) vp.second = ++cnt;
    for (int i = 1; i <= n; i++)
      for (auto &v : q[i]) v.first = mp[v.first], v.second = mp[v.second];
    for (int i = 1; i <= n; i++) {
      pair<int, int> tmp = {0, 0};
      for (auto lr : q[i]) tmp = max(tmp, query(lr.first, lr.second));
      dp[i] = tmp.first + 1, tr[i] = tmp.second;
      for (auto lr : q[i]) update(lr.first, lr.second, {dp[i], i});
    }
    int pos = max_element(dp + 1, dp + n + 1) - dp;
    while (pos) {
      chk[pos] = 1, pos = tr[pos];
    }
    vector<int> res;
    for (int i = 1; i <= n; i++)
      if (!chk[i]) res.push_back(i);
    cout << res.size() << '\n';
    for (auto v : res) cout << v << ' ';
  }
  