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;
  const int N = 3e5 + 10;
  const int M = 1e9 + 7;
  int n, m, ind, l, r;
  int prv[N], a[2 * N];
  struct Interv {
    int ind, l, r;
  } interv[N];
  pair<int, int> arb[8 * N], lazy[8 * N];
  bitset<N> mark;
  void push(int node, int l, int r) {
    if (l == r || !lazy[node].first) return;
    lazy[node * 2] = arb[node * 2] = lazy[node];
    lazy[node * 2 + 1] = arb[node * 2 + 1] = lazy[node];
    lazy[node] = {0, 0};
  }
  void setValue(int node, int l, int r, int L, int R, pair<int, int> value) {
    if (L > R) return;
    if (L <= l && r <= R) {
      arb[node] = lazy[node] = value;
      return;
    }
    int mid = (l + r) / 2;
    push(node, l, r);
    setValue(node * 2, l, mid, L, min(R, mid), value);
    setValue(node * 2 + 1, mid + 1, r, max(L, mid + 1), R, value);
    arb[node] = max(arb[node * 2], arb[node * 2 + 1]);
  }
  pair<int, int> query(int node, int l, int r, int L, int R) {
    if (L > R) return {-1, 0};
    if (L <= l && r <= R) return arb[node];
    push(node, l, r);
    int mid = (l + r) / 2;
    return max(query(node * 2, l, mid, L, min(R, mid)),
               query(node * 2 + 1, mid + 1, r, max(L, mid + 1), R));
  }
  bool inline cmp(Interv a, Interv b) { return a.ind < b.ind; }
  signed main() {
    int t = 0, i, j, row;
    cin >> n >> m;
    for (i = 1; i <= m; i++) {
      cin >> ind >> l >> r;
      interv[i] = {ind, l, r};
      a[++t] = l;
      a[++t] = r;
    }
    sort(a + 1, a + t + 1);
    int y = 0;
    for (i = 2; i <= t + 1; i++)
      if (a[i] != a[i - 1]) a[++y] = a[i - 1];
    t = y;
    for (i = 1; i <= m; i++) {
      interv[i].l = lower_bound(a + 1, a + t + 1, interv[i].l) - a;
      interv[i].r = lower_bound(a + 1, a + t + 1, interv[i].r) - a;
    }
    sort(interv + 1, interv + m + 1, cmp);
    pair<int, int> ans = {0, 0};
    pair<int, int> mx = {0, 0};
    for (row = 1; row <= m + 1; row++) {
      if (interv[row].ind != interv[row - 1].ind && row > 1) {
        for (i = row - 1; interv[i].ind == interv[row - 1].ind; i--)
          setValue(1, 1, 2 * m, interv[i].l, interv[i].r,
                   {mx.first + 1, interv[i].ind});
        prv[interv[row - 1].ind] = mx.second;
        ans = max(ans, {mx.first + 1, interv[row - 1].ind});
        mx = {0, 0};
      }
      if (row <= m)
        mx = max(mx, query(1, 1, 2 * m, interv[row].l, interv[row].r));
    }
    int choose = 0;
    row = ans.second;
    do {
      ++choose;
      mark[row] = 1;
      row = prv[row];
    } while (row);
    cout << n - choose << '\n';
    for (i = 1; i <= n; i++)
      if (!mark[i]) cout << i << ' ';
    return 0;
  }
  