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;
  struct tri {
    long long x, y, z;
    tri() : x(0), y(0), z(0) {}
    tri(const long long x, const long long y, const long long z)
        : x(x), y(y), z(z) {}
    tri(const tri &t) : x(t.x), y(t.y), z(t.z) {}
  };
  bool query_compare(const tri &a, const tri &b) { return a.x < b.x; }
  class OverlapCompare {
   public:
    bool operator()(const tri &a, const tri &b) {
      if (a.y == b.y) return a.x < b.x;
      return a.y < b.y;
    }
  };
  long long n, m;
  tri querys[300005];
  set<tri, OverlapCompare> overlap;
  long long dp[300005];
  long long record[300005];
  list<tri> be_add, be_remove;
  bool removed[300005];
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> m;
    for (long long i = 0; i < m; i++) {
      cin >> querys[i].x >> querys[i].y >> querys[i].z;
    }
    stable_sort(querys, querys + m, query_compare);
    fill_n(dp, n + 1, 1);
    for (long long i = 0; i < m; i++) {
      set<tri, OverlapCompare>::iterator lb =
          overlap.lower_bound(tri(-1LL, querys[i].y, querys[i].z));
      set<tri, OverlapCompare>::iterator ub =
          overlap.upper_bound(tri((long long)1e15, querys[i].z, 0LL));
      if (lb != overlap.begin()) {
        if ((--lb)->z < querys[i].y) {
          ++lb;
        }
      }
      for (set<tri, OverlapCompare>::iterator it = lb; it != ub; it++) {
        if (it->x == querys[i].x) continue;
        if (dp[querys[i].x] < dp[it->x] + 1) {
          dp[querys[i].x] = dp[it->x] + 1;
          record[querys[i].x] = it->x;
        }
      }
      be_add.clear();
      be_remove.clear();
      for (set<tri, OverlapCompare>::iterator it = lb; it != ub; it++) {
        bool cur_be_remove = false;
        if (it->y < querys[i].y && it->z >= querys[i].y) {
          be_add.push_back(tri(it->x, it->y, querys[i].y - 1));
          cur_be_remove = true;
        }
        if (it->y <= querys[i].z && querys[i].z < it->z) {
          be_add.push_back(tri(it->x, querys[i].z + 1, it->z));
          cur_be_remove = true;
        }
        if (it->y >= querys[i].y && it->z <= querys[i].z) {
          cur_be_remove = true;
        }
        if (cur_be_remove) {
          be_remove.push_back(*it);
        }
      }
      for (list<tri>::iterator it = be_remove.begin(); it != be_remove.end();
           it++) {
        overlap.erase(*it);
      }
      for (list<tri>::iterator it = be_add.begin(); it != be_add.end(); it++) {
        overlap.insert(*it);
      }
      overlap.insert(querys[i]);
    }
    long long rst_idx = 1;
    for (long long i = 1; i <= n; i++) {
      if (dp[i] > dp[rst_idx]) rst_idx = i;
    }
    long long i = rst_idx;
    fill_n(removed, n + 1, true);
    while (i != 0) {
      removed[i] = false;
      i = record[i];
    }
    cout << n - dp[rst_idx] << '\n';
    for (long long i = 1; i <= n; i++) {
      if (removed[i]) cout << i << ' ';
    }
    cout << '\n';
    return 0;
  }
  