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")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
  using namespace std;
  const double PI = 3.141592653589793238463;
  const int INF = 0x3f3f3f3f;
  const int INFS = 1000000001;
  const long long LLINF = 9223372036854775807;
  const long long M = 1000000007;
  const double EPS = 1e-8;
  struct node {
    int l, r;
    pair<int, int> mn = {1000000, -1}, d = {1000000, -1};
  };
  vector<node> st;
  void build(int v, int l, int r) {
    st[v].l = l;
    st[v].r = r;
    if (l == r) return;
    int mid = (l + r) / 2;
    build(2 * v + 1, l, mid);
    build(2 * v + 2, mid + 1, r);
  }
  void push(int v) {
    st[v].mn = min(st[v].mn, st[v].d);
    if (st[v].l != st[v].r) {
      st[2 * v + 1].d = min(st[2 * v + 1].d, st[v].d);
      st[2 * v + 2].d = min(st[2 * v + 2].d, st[v].d);
    }
    st[v].d = {1000000, -1};
  }
  pair<int, int> get(int v, int l, int r) {
    push(v);
    if (l > st[v].r || r < st[v].l) return {1000000, -1};
    if (l <= st[v].l && r >= st[v].r) return st[v].mn;
    return min(get(2 * v + 1, l, r), get(2 * v + 2, l, r));
  }
  void upd(int v, int l, int r, int val, int idx) {
    push(v);
    if (l > st[v].r || r < st[v].l) return;
    if (l <= st[v].l && r >= st[v].r) {
      st[v].d = {val, idx};
      push(v);
      return;
    }
    upd(2 * v + 1, l, r, val, idx);
    upd(2 * v + 2, l, r, val, idx);
    st[v].mn = min(st[2 * v + 1].mn, st[2 * v + 2].mn);
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
    int n, q;
    cin >> n >> q;
    vector<vector<pair<int, int> > > segs(n);
    int t, l, r;
    map<int, int> id = {};
    for (int i = 0; i < q; i++) {
      cin >> t >> l >> r;
      --t;
      segs[t].push_back({l, r});
      id[l] = 0;
      id[r] = 0;
    }
    int m = 0;
    for (auto x : id) id[x.first] = m++;
    st.resize(4 * m);
    build(0, 0, m - 1);
    vector<int> dp(n), pr(n);
    dp[0] = n - 1;
    pr[0] = 0;
    for (auto seg : segs[0]) upd(0, id[seg.first], id[seg.second], dp[0], 0);
    pair<int, int> got;
    for (int i = 1; i < n; i++) {
      dp[i] = n - 1;
      pr[i] = i;
      for (auto seg : segs[i]) {
        got = get(0, id[seg.first], id[seg.second]);
        if (got.first - 1 < dp[i]) {
          dp[i] = got.first - 1;
          pr[i] = got.second;
        }
      }
      for (auto seg : segs[i]) upd(0, id[seg.first], id[seg.second], dp[i], i);
    }
    pair<int, int> mn = {dp[0], 0};
    for (int i = 1; i < n; i++) {
      if (dp[i] < mn.first) {
        mn.first = dp[i];
        mn.second = i;
      }
    }
    vector<int> ans = {};
    vector<bool> used(n, false);
    int cr = mn.second;
    while (cr != pr[cr]) {
      used[cr] = true;
      cr = pr[cr];
    }
    used[cr] = true;
    for (int i = 0; i < n; i++) {
      if (!used[i]) ans.push_back(i + 1);
    }
    if (ans.size() != mn.first) return -1;
    cout << ans.size() << "\n";
    for (auto x : ans) cout << x << ' ';
    cout << "\n";
    return 0;
  }
  