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 int MAXN = 3e5 + 7;
  const int INF = 1e9 + 7;
  struct segment_tree {
    vector<pair<int, int>> values;
    vector<pair<int, int>> add;
    int BASE;
    segment_tree(int size) {
      BASE = 1;
      while (BASE <= size) BASE <<= 1;
      values.resize(2 * BASE);
      add.resize(2 * BASE);
    }
    pair<int, int> best(pair<int, int> lhs, pair<int, int> rhs) {
      if (lhs.first < rhs.first) return lhs;
      return rhs;
    }
    void push(int v) {
      int l = 2 * v;
      int r = 2 * v + 1;
      values[l] = best(values[l], add[v]);
      values[r] = best(values[r], add[v]);
      add[l] = best(add[l], add[v]);
      add[r] = best(add[r], add[v]);
      add[v] = {INF, -1};
    }
    pair<int, int> minn(int l, int r) { return minn(l, r, 1, 0, BASE - 1); }
    pair<int, int> minn(int l, int r, int v, int b, int e) {
      if (r < b || e < l) return {INF, -1};
      if (l <= b && e <= r) return values[v];
      push(v);
      int mid = (b + e) / 2;
      pair<int, int> lVal = minn(l, r, 2 * v, b, mid);
      pair<int, int> rVal = minn(l, r, 2 * v + 1, mid + 1, e);
      values[v] = best(values[2 * v], values[2 * v + 1]);
      return best(lVal, rVal);
    }
    void change(int l, int r, pair<int, int> val) {
      change(l, r, val, 1, 0, BASE - 1);
    }
    void change(int l, int r, pair<int, int> val, int v, int b, int e) {
      if (r < b || e < l) return;
      if (l <= b && e <= r) {
        add[v] = val;
        values[v] = val;
        return;
      }
      push(v);
      int mid = (b + e) / 2;
      change(l, r, val, 2 * v, b, mid);
      change(l, r, val, 2 * v + 1, mid + 1, e);
      values[v] = best(values[2 * v], values[2 * v + 1]);
    }
  };
  vector<pair<int, int>> segments[MAXN];
  int dp[MAXN];
  int prv[MAXN];
  bool ok[MAXN];
  void solve() {
    int n, m;
    cin >> n >> m;
    map<int, int> scaled;
    vector<int> sorted;
    for (int i = 1; i <= m; ++i) {
      int h, l, r;
      cin >> h >> l >> r;
      segments[h].push_back({l, r});
      sorted.push_back(l);
      sorted.push_back(r);
    }
    sort(sorted.begin(), sorted.end());
    int last = -1;
    int maxVal = 0;
    for (int i = 0; i < sorted.size(); ++i) {
      if (sorted[i] != last) {
        scaled[sorted[i]] = ++maxVal;
        last = sorted[i];
      }
    }
    for (int h = 1; h <= n; ++h) {
      for (auto& [l, r] : segments[h]) {
        l = scaled[l];
        r = scaled[r];
      }
    }
    segment_tree tree(maxVal);
    for (int h = 1; h <= n; ++h) {
      dp[h] = INF;
      for (auto [l, r] : segments[h]) {
        auto [val, ind] = tree.minn(l, r);
        if (val + h - 1 < dp[h]) {
          dp[h] = val + h - 1;
          prv[h] = ind;
        }
      }
      for (auto [l, r] : segments[h]) {
        tree.change(l, r, {dp[h] - h, h});
      }
    }
    auto [ans, ind] = tree.minn(1, maxVal);
    cout << ans + n << '\n';
    while (ind != 0) {
      ok[ind] = true;
      ind = prv[ind];
    }
    for (int i = 1; i <= n; ++i) {
      if (!ok[i]) cout << i << ' ';
    }
    cout << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(0), cin.tie(0);
    int T;
    T = 1;
    while (T--) solve();
    return 0;
  }
  