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 double eps = 1e-10, pi = 3.1415926535898;
  const int mod = 1e9 + 7, maxn = 3e5 + 10;
  int T, n, m, k, x, y, a[maxn], dp[maxn];
  struct P {
    int x, y, i;
    bool operator<(const P& a) const {
      if (x == a.x) return y < a.y;
      return x < a.x;
    }
  } p[2 * maxn];
  vector<int> v[maxn], v1[maxn];
  int main(int argc, char const* argv[]) {
    int ans = 0, cnt = 0;
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; ++i) {
      dp[i] = 1;
    }
    set<int> s;
    for (int i = 1; i <= m; ++i) {
      scanf("%d %d %d", &k, &x, &y);
      p[2 * i - 1].x = x;
      p[2 * i - 1].y = 0;
      p[2 * i - 1].i = k;
      p[2 * i].x = y;
      p[2 * i].y = 1;
      p[2 * i].i = k;
    }
    sort(p + 1, p + 2 * m + 1);
    for (int i = 1; i <= 2 * m; ++i) {
      if (p[i].y == 0) {
        ++a[p[i].i];
        if (a[p[i].i] == 1) {
          set<int>::iterator it = s.lower_bound(p[i].i);
          if (it != s.end()) {
            v[p[i].i].push_back(*it);
            v1[*it].push_back(p[i].i);
          }
          if (!s.empty() and it != s.begin()) {
            --it;
            v[*it].push_back(p[i].i);
            v1[p[i].i].push_back(*it);
          }
          s.insert(p[i].i);
        }
      } else {
        --a[p[i].i];
        if (a[p[i].i] == 0) {
          s.erase(p[i].i);
        }
      }
    }
    for (int i = 1; i <= n; ++i) {
      for (int j : v[i]) {
        dp[j] = max(dp[j], dp[i] + 1);
        if (dp[j] > ans) {
          ans = dp[j];
          cnt = j;
        }
      }
    }
    a[cnt] = mod;
    while (dp[cnt] > 1) {
      for (int i : v1[cnt]) {
        if (dp[i] == dp[cnt] - 1) {
          a[i] = mod;
          cnt = i;
          break;
        }
      }
    }
    printf("%d\n", n - ans);
    for (int i = 1; i <= n; ++i) {
      if (a[i] != mod) printf("%d ", i);
    }
    printf("\n");
    return 0;
  }
  