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;
  pair<int, int> seg[2100000], lazy[2100000];
  pair<int, int> Merge(pair<int, int> l, pair<int, int> r) {
    if (l < r)
      return l;
    else
      return r;
  }
  void Propagate(int n, int l, int r) {
    if (lazy[n] != make_pair(0x3f3f3f3f, 0)) {
      if (l != r) {
        lazy[n << 1] = Merge(lazy[n << 1], lazy[n]);
        lazy[n << 1 | 1] = Merge(lazy[n << 1 | 1], lazy[n]);
      }
      seg[n] = Merge(seg[n], lazy[n]);
      lazy[n] = make_pair(0x3f3f3f3f, 0);
    }
  }
  pair<int, int> Update(int L, int R, pair<int, int> val, int n, int l, int r) {
    Propagate(n, l, r);
    if (r < L || R < l) return seg[n];
    if (L <= l && r <= R) {
      lazy[n] = Merge(lazy[n], val);
      Propagate(n, l, r);
      return seg[n];
    }
    int mid = l + r >> 1;
    return seg[n] = Merge(Update(L, R, val, n << 1, l, mid),
                          Update(L, R, val, n << 1 | 1, mid + 1, r));
  }
  pair<int, int> Query(int L, int R, int n, int l, int r) {
    Propagate(n, l, r);
    if (r < L || R < l) return make_pair(0x3f3f3f3f, 0);
    if (L <= l && r <= R) return seg[n];
    int mid = l + r >> 1;
    return Merge(Query(L, R, n << 1, l, mid),
                 Query(L, R, n << 1 | 1, mid + 1, r));
  }
  int N, M;
  vector<pair<int, int> > v[300001];
  pair<int, int> dp[300001];
  int main() {
    for (int i = 0; i < 2100000; i++) seg[i] = lazy[i] = make_pair(0x3f3f3f3f, 0);
    ios::sync_with_stdio(0);
    cin.tie(0);
    cin.exceptions(ios::badbit | ios::failbit);
    cin >> N >> M;
    vector<int> com;
    for (int i = 1; i <= M; i++) {
      int n, l, r;
      cin >> n >> l >> r;
      r++;
      v[n].push_back({l, r});
      com.push_back(l);
      com.push_back(r);
    }
    sort(com.begin(), com.end());
    com.erase(unique(com.begin(), com.end()), com.end());
    for (int i = 1; i <= N; i++)
      for (auto &[l, r] : v[i]) {
        l = lower_bound(com.begin(), com.end(), l) - com.begin();
        r = lower_bound(com.begin(), com.end(), r) - com.begin();
      }
    for (int i = 1; i <= N; i++) {
      dp[i] = {i - 1, 0};
      for (auto [l, r] : v[i]) {
        auto it = Query(l, r - 1, 1, 0, (int)com.size() - 1);
        dp[i] = min(dp[i], make_pair(it.first + i, it.second));
      }
      for (auto [l, r] : v[i]) {
        Update(l, r - 1, make_pair(dp[i].first - i - 1, i), 1, 0,
               (int)com.size() - 1);
      }
    }
    int res = 0x3f3f3f3f, n = 0;
    for (int i = 1; i <= N; i++) {
      int val = dp[i].first + (N - i);
      if (res > val) res = val, n = i;
    }
    set<int> s;
    for (int i = 1; i <= N; i++) s.insert(i);
    while (n) {
      s.erase(n);
      n = dp[n].second;
    }
    cout << res << "\n";
    if (res) {
      for (int i : s) cout << i << " ";
      cout << "\n";
    }
    return 0;
  }
  