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("O2")
  #pragma gcc optimize("unroll-loops")
  using namespace std;
  using ll = long long;
  using pii = pair<int, int>;
  const int N = 6e5 + 10;
  int n, m;
  vector<vector<int>> inp;
  map<int, int> conv;
  int mx;
  vector<pii> vec[N];
  pii t[4 * N];
  pii lazy[4 * N];
  void push(int v) {
    lazy[v * 2] = max(lazy[v * 2], lazy[v]);
    lazy[v * 2 + 1] = max(lazy[v * 2 + 1], lazy[v]);
    t[v * 2] = max(t[v * 2], lazy[v]);
    t[v * 2 + 1] = max(t[v * 2 + 1], lazy[v]);
  }
  void update(int v, int tl, int tr, int l, int r, pii val) {
    if (l > r || tr < l || r < tl) return;
    if (l <= tl && tr <= r) {
      t[v] = max(t[v], val);
      lazy[v] = max(lazy[v], val);
      return;
    }
    int tm = (tl + tr) / 2;
    push(v);
    update(v * 2, tl, tm, l, r, val);
    update(v * 2 + 1, tm + 1, tr, l, r, val);
    t[v] = max(t[v * 2], t[v * 2 + 1]);
  }
  pii ask(int v, int tl, int tr, int l, int r) {
    if (l > r || tr < l || r < tl) return make_pair(0, 0);
    if (l <= tl && tr <= r) return t[v];
    int tm = (tl + tr) / 2;
    push(v);
    return max(ask(v * 2, tl, tm, l, r), ask(v * 2 + 1, tm + 1, tr, l, r));
  }
  int dp[N], par[N];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m;
    inp.resize(m);
    for (int i = 0; i < m; i++) {
      for (int j = 0; j < 3; j++) {
        int a;
        cin >> a;
        inp[i].push_back(a);
        if (j) conv[a] = 1;
      }
    }
    {
      mx = 0;
      for (auto& x : conv) {
        x.second = ++mx;
      }
      for (auto& x : inp) {
        for (int i = 1; i < 3; i++) {
          x[i] = conv[x[i]];
        }
        vec[x[0]].push_back(make_pair(x[1], x[2]));
      }
    }
    int ans = 0;
    int cur = 0;
    for (int i = 1; i <= n; i++) {
      pii p = make_pair(0, 0);
      for (auto& x : vec[i]) {
        p = max(p, ask(1, 1, mx, x.first, x.second));
      }
      dp[i] = p.first + 1;
      par[i] = p.second;
      for (auto& x : vec[i]) {
        update(1, 1, mx, x.first, x.second, make_pair(dp[i], i));
      }
      if (dp[i] > dp[cur]) cur = i;
    }
    vector<bool> msk(n + 1, 0);
    int len = n;
    while (cur != 0) {
      msk[cur] = 1;
      cur = par[cur];
      len--;
    }
    cout << len << '\n';
    if (len) {
      for (int i = 1; i <= n; i++) {
        if (msk[i]) continue;
        cout << i << ' ';
      }
      cout << '\n';
    }
    return 0;
  }
  