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;
  vector<pair<int, int> > vec[2102100];
  pair<int, int> tr[2102100];
  int lz[2102100];
  int id[2102100];
  int l[2102100];
  int r[2102200];
  int b[2102100];
  int path[2102100];
  int st[2102100];
  void build(int p, int l, int r) {
    tr[p] = {0, -1};
    lz[p] = 0;
    if (l == r) {
      return;
    }
    int mid = l + r >> 1;
    build(p << 1, l, mid);
    build(p << 1 | 1, mid + 1, r);
  }
  void pushup(int p) { tr[p] = max(tr[p << 1], tr[p << 1 | 1]); }
  void pushdown(int p) {
    if (lz[p]) {
      lz[p << 1] = lz[p << 1 | 1] = lz[p];
      lz[p] = 0;
      tr[p << 1] = tr[p << 1 | 1] = tr[p];
    }
  }
  pair<int, int> query(int p, int l, int r, int ll, int rr) {
    if (ll <= l && rr >= r) {
      return tr[p];
    }
    pushdown(p);
    int mid = l + r >> 1;
    pair<int, int> res = {0, -1};
    if (ll <= mid) res = max(query(p << 1, l, mid, ll, rr), res);
    if (rr > mid) res = max(query(p << 1 | 1, mid + 1, r, ll, rr), res);
    return res;
  }
  void update(int p, int l, int r, int ll, int rr, pair<int, int> res) {
    if (ll <= l && rr >= r) {
      tr[p] = res;
      lz[p] = 1;
      return;
    }
    pushdown(p);
    int mid = l + r >> 1;
    if (ll <= mid) update(p << 1, l, mid, ll, rr, res);
    if (rr > mid) update(p << 1 | 1, mid + 1, r, ll, rr, res);
    pushup(p);
  }
  int main() {
    int n, t;
    cin >> n >> t;
    for (int i = 1; i <= t; i++) {
      cin >> id[i] >> l[i] >> r[i];
      b[i * 2 - 1] = l[i];
      b[i * 2] = r[i];
    }
    sort(b + 1, b + 1 + t * 2);
    int len = unique(b + 1, b + 1 + t * 2) - b - 1;
    build(1, 1, len);
    for (int i = 1; i <= t; i++) {
      l[i] = lower_bound(b + 1, b + 1 + len, l[i]) - b;
      r[i] = lower_bound(b + 1, b + 1 + len, r[i]) - b;
      vec[id[i]].push_back({l[i], r[i]});
    }
    for (int i = 1; i <= n; i++) {
      pair<int, int> mx = {0, -1};
      for (auto tt : vec[i]) {
        mx = max(mx, query(1, 1, len, tt.first, tt.second));
      }
      mx.first++;
      path[i] = mx.second;
      mx.second = i;
      for (auto tt : vec[i]) {
        update(1, 1, len, tt.first, tt.second, mx);
      }
    }
    pair<int, int> ans = query(1, 1, len, 1, len);
    int u = ans.second;
    int aa = n;
    while (u != -1) {
      st[u] = 1;
      aa--;
      u = path[u];
    }
    cout << aa << endl;
    for (int i = 1; i <= n; i++) {
      if (!st[i]) cout << i << " ";
    }
  }
  