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>
  #pragma GCC optimize("Ofast")
  using namespace std;
  const int N = 2e6 + 10;
  const long long mod = 1e9 + 7;
  const long long inf = 8e18;
  const int LOG = 20;
  long long pw(long long a, long long b, long long M = mod, long long ret = 1) {
    while (b) {
      ret = ret * (b & 1 ? a : 1) % M, a = a * a % M, b >>= 1;
    }
    return ret;
  }
  int n, m, ptr, sz, last[N], dp[N], in[N];
  vector<pair<int, int> > vec[N];
  vector<int> cm;
  map<int, int> mp;
  pair<int, int> seg[N << 2], lz[N << 2];
  void upd(int l, int r, pair<int, int> x, int v = 1, int tl = 1, int tr = ptr) {
    if (l > tr || r < tl || l > r) return;
    if (l <= tl && tr <= r) {
      seg[v] = max(seg[v], x);
      lz[v] = max(lz[v], x);
      return;
    }
    int mid = (tl + tr) >> 1;
    upd(l, r, x, v << 1, tl, mid), upd(l, r, x, v << 1 | 1, mid + 1, tr);
    seg[v] = max(seg[v << 1], seg[v << 1 | 1]);
  }
  pair<int, int> get(int l, int r, int v = 1, int tl = 1, int tr = ptr) {
    if (l > tr || r < tl || l > r) return make_pair(0, 0);
    if (l <= tl && tr <= r) {
      return seg[v];
    }
    int mid = (tl + tr) >> 1;
    return max(
        {lz[v], get(l, r, v << 1, tl, mid), get(l, r, v << 1 | 1, mid + 1, tr)});
  }
  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);
      cm.push_back(l);
      cm.push_back(r);
      vec[id].push_back(make_pair(l, r));
    }
    cm.push_back(-1e9), cm.push_back(1e9);
    sort(cm.begin(), cm.end());
    cm.resize(unique(cm.begin(), cm.end()) - cm.begin());
    sz = (int)cm.size();
    for (int i = 0; i < sz; i++) {
      mp[cm[i]] = ++ptr;
      if (i + 1 != sz && cm[i] + 1 < cm[i + 1]) {
        ptr++;
      }
    }
    int Mx = 0;
    for (int i = 1; i <= n; i++) {
      dp[i] = 1;
      for (auto cu : vec[i]) {
        pair<int, int> calc = get(mp[cu.first], mp[cu.second]);
        if (calc.first + 1 > dp[i]) {
          dp[i] = calc.first + 1;
          last[i] = calc.second;
        }
      }
      for (auto cu : vec[i]) {
        upd(mp[cu.first], mp[cu.second], make_pair(dp[i], i));
      }
      if (dp[i] > dp[Mx]) Mx = i;
    }
    while (Mx > 0) {
      in[Mx] = 1;
      Mx = last[Mx];
    }
    vector<int> Ans;
    for (int i = 1; i <= n; i++) {
      if (in[i] == 0) Ans.push_back(i);
    }
    printf("%d\n", (int)Ans.size());
    for (auto x : Ans) {
      printf("%d ", x);
    }
    return 0;
  }
  