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 maxn = 1e6 + 7;
  const int inf = 2e9;
  const long long inff = 1e18;
  const long long mod = 1e9 + 7;
  const long long C = 1e6;
  struct segtree {
    long long seg[maxn * 4], tag[maxn * 4];
    void init() {
      memset(seg, 0, sizeof(seg));
      memset(tag, 0, sizeof(tag));
    }
    void push(long long now, long long l, long long r) {
      if (tag[now] == 0) return;
      tag[now * 2] = tag[now * 2 + 1] = seg[now * 2] = seg[now * 2 + 1] =
          tag[now];
      tag[now] = 0;
    }
    void modify(long long now, long long l, long long r, long long ql,
                long long qr, long long v) {
      if (l >= ql && r <= qr) {
        tag[now] = seg[now] = v;
        return;
      }
      push(now, l, r);
      long long m = (l + r) >> 1;
      if (ql <= m) modify(now * 2, l, m, ql, qr, v);
      if (qr >= m + 1) modify(now * 2 + 1, m + 1, r, ql, qr, v);
      seg[now] = max(seg[now * 2], seg[now * 2 + 1]);
    }
    long long query(long long now, long long l, long long r, long long ql,
                    long long qr) {
      if (l >= ql && r <= qr) return seg[now];
      push(now, l, r);
      long long m = (l + r) >> 1;
      if (qr <= m) return query(now * 2, l, m, ql, qr);
      if (ql > m) return query(now * 2 + 1, m + 1, r, ql, qr);
      return max(query(now * 2, l, m, ql, qr),
                 query(now * 2 + 1, m + 1, r, ql, qr));
    }
  } tree;
  long long dp[maxn], prevv[maxn];
  vector<pair<long long, long long> > v[maxn];
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    long long n, m;
    cin >> n >> m;
    vector<long long> values;
    for (long long i = 1; i <= m; i++) {
      long long x, y, z;
      cin >> x >> y >> z;
      v[x].push_back({y, z});
      values.push_back(y);
      values.push_back(z);
    }
    sort((values).begin(), (values).end());
    values.resize(unique((values).begin(), (values).end()) - values.begin());
    long long N = ((long long)(values).size());
    for (long long i = 1; i <= n; i++) {
      for (pair<long long, long long> &j : v[i]) {
        j.first = lower_bound((values).begin(), (values).end(), j.first) -
                  values.begin() + 1;
        j.second = lower_bound((values).begin(), (values).end(), j.second) -
                   values.begin() + 1;
      }
    }
    tree.init();
    pair<long long, long long> maxx = {-1, -1};
    for (long long i = 1; i <= n; i++) {
      long long mx = 0;
      for (pair<long long, long long> j : v[i])
        mx = max(mx, tree.query(1, 1, N, j.first, j.second));
      prevv[i] = mx % C;
      dp[i] = (mx / C + 1) * C + i;
      for (pair<long long, long long> j : v[i])
        tree.modify(1, 1, N, j.first, j.second, dp[i]);
      maxx = max(maxx, {dp[i], i});
    }
    vector<long long> used(n), ans;
    long long ttmp = maxx.second;
    while (ttmp) {
      used[ttmp - 1] = 1;
      ttmp = prevv[ttmp];
    }
    for (long long i = 0; i < n; i++) {
      if (!used[i]) ans.push_back(i + 1);
    }
    cout << ((long long)(ans).size()) << '\n';
    for (long long i : ans) cout << i << ' ';
    cout << '\n';
  }
  