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 NMAX = 1000001;
  map<int, int> used;
  vector<vector<pair<int, int> > > seg(NMAX);
  pair<int, int> t[8 * NMAX];
  pair<int, int> lazy[8 * NMAX];
  void push(int v) {
    if (lazy[v].first == 1e9 + 1) return;
    t[v * 2] = lazy[v];
    t[2 * v + 1] = lazy[v];
    lazy[2 * v] = lazy[v];
    lazy[2 * v + 1] = lazy[v];
    lazy[v] = {1e9 + 1, 1e9 + 1};
  }
  pair<int, int> get(int v, int l, int r, int st, int fin) {
    if (l > fin || r < st) return {-1e9 + 1, -1e9 + 1};
    if (l >= st && r <= fin) {
      return t[v];
    }
    push(v);
    int m = (l + r) / 2;
    pair<int, int> x = get(2 * v, l, m, st, fin);
    pair<int, int> y = get(2 * v + 1, m + 1, r, st, fin);
    return max(x, y);
  }
  void update(int v, int l, int r, int st, int fin, int nw, int idx) {
    if (l > fin || st > r) return;
    if (l >= st && r <= fin) {
      t[v] = {nw, idx};
      lazy[v] = {nw, idx};
      return;
    }
    push(v);
    int m = (l + r) / 2;
    update(2 * v, l, m, st, fin, nw, idx);
    update(2 * v + 1, m + 1, r, st, fin, nw, idx);
    t[v] = max(t[2 * v], t[2 * v + 1]);
  }
  map<int, int> mp;
  int R = 2 * 1e6;
  void solve() {
    int n, m;
    cin >> n >> m;
    vector<int> vec;
    for (int i = 1; i <= m; i++) {
      int pos, l, r;
      cin >> pos >> l >> r;
      seg[pos].emplace_back(l, r);
      vec.push_back(l);
      vec.push_back(r);
    }
    sort(vec.begin(), vec.end());
    for (int i = 0; i < vec.size(); i++) {
      mp[vec[i]] = i + 1;
    }
    int dp[n + 1], par[n + 1];
    dp[0] = 0;
    update(1, 1, R, 1, R, 0, 0);
    for (int i = 1; i <= n; i++) {
      pair<int, int> ans = {-1e9 + 1, -1e9 + 1};
      for (int j = 0; j < seg[i].size(); j++) {
        int l = mp[seg[i][j].first], r = mp[seg[i][j].second];
        pair<int, int> x = get(1, 1, R, l, r);
        ans = max(ans, x);
      }
      dp[i] = ans.first + 1;
      par[i] = ans.second;
      for (int j = 0; j < seg[i].size(); j++) {
        int l = mp[seg[i][j].first], r = mp[seg[i][j].second];
        update(1, 1, R, l, r, dp[i], i);
      }
    }
    int mx = -1;
    int mxi;
    for (int i = 1; i <= n; i++) {
      if (mx < dp[i]) {
        mx = dp[i];
        mxi = i;
      }
    }
    set<int> second;
    for (int i = 1; i <= n; i++) second.insert(i);
    cout << n - mx << "\n";
    while (mxi) {
      second.erase(mxi);
      mxi = par[mxi];
    }
    for (int to : second) cout << to << ' ';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int test = 1;
    while (test--) {
      solve();
    }
  }
  