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;
  int n, m, tot;
  pair<int, int> data[300010 << 3], tag[300010 << 3];
  int a[600010], nn, pre[300010], ans, ansp, used[300010];
  vector<pair<int, int> > G[300010];
  pair<int, int> piimax(pair<int, int> x, pair<int, int> y) {
    return x.first > y.first ? x : y;
  }
  void push_down(int cur) {
    if (tag[cur].first == 0) return;
    data[cur << 1] = data[cur << 1 | 1] = tag[cur];
    tag[cur << 1] = tag[cur << 1 | 1] = tag[cur];
    tag[cur] = make_pair(0, 0);
  }
  void push_up(int cur) {
    data[cur] = piimax(data[cur << 1], data[cur << 1 | 1]);
  }
  void update(int cur, int l, int r, int L, int R, pair<int, int> v) {
    if (L <= l && r <= R) {
      data[cur] = tag[cur] = v;
      return;
    }
    push_down(cur);
    int mid = (l + r) >> 1;
    if (L <= mid) update(cur << 1, l, mid, L, R, v);
    if (mid < R) update(cur << 1 | 1, mid + 1, r, L, R, v);
    push_up(cur);
  }
  pair<int, int> query(int cur, int l, int r, int L, int R) {
    if (L <= l && r <= R) return data[cur];
    push_down(cur);
    int mid = (l + r) >> 1;
    if (L <= mid && mid < R)
      return piimax(query(cur << 1, l, mid, L, R),
                    query(cur << 1 | 1, mid + 1, r, L, R));
    else if (L <= mid)
      return query(cur << 1, l, mid, L, R);
    else
      return query(cur << 1 | 1, mid + 1, r, L, R);
  }
  int main() {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= m; ++i) {
      int id, l, r;
      scanf("%d %d %d", &id, &l, &r);
      G[id].push_back(make_pair(l, r));
      a[tot++] = l;
      a[tot++] = r;
    }
    sort(a, a + tot);
    nn = unique(a, a + tot) - a;
    for (int i = 1; i <= n; ++i) {
      int gi = G[i].size();
      for (int j = 0; j < gi; ++j) {
        G[i][j].first = lower_bound(a, a + nn, G[i][j].first) - a + 1;
        G[i][j].second = lower_bound(a, a + nn, G[i][j].second) - a + 1;
      }
    }
    for (int i = 1; i <= n; ++i) {
      int gi = G[i].size();
      int mx = -1;
      for (int j = 0; j < gi; ++j) {
        pair<int, int> t = query(1, 1, nn, G[i][j].first, G[i][j].second);
        if (t.first > mx) {
          mx = t.first;
          pre[i] = t.second;
        }
      }
      if (mx + 1 > ans) {
        ans = mx + 1;
        ansp = i;
      }
      for (int j = 0; j < gi; ++j) {
        pair<int, int> tt = make_pair(mx + 1, i);
        update(1, 1, nn, G[i][j].first, G[i][j].second, tt);
      }
    }
    printf("%d\n", n - ans);
    for (int i = ansp; i; i = pre[i]) used[i] = 1;
    for (int i = 1; i <= n; ++i)
      if (!used[i]) printf("%d ", i);
    return 0;
  }
  