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 N = 6e5;
  const int M = N + 7;
  const double EPS = 1e-10;
  const int MOD = 1e9 + 7;
  const long long INF = 1e18 + 17;
  int n, m, dp[M];
  vector<pair<int, int> > g[M];
  int t[4 * M], add[4 * M];
  void push(int v, int tl, int tr) {
    int tm = (tl + tr) >> 1;
    if (add[v] != -1) {
      t[v + v] = add[v];
      add[v + v] = add[v];
      t[v + v + 1] = add[v];
      add[v + v + 1] = add[v];
      add[v] = -1;
    }
  }
  void upd(int v, int tl, int tr, int l, int r, int x) {
    if (l > r) return;
    if (tl == l && tr == r) {
      t[v] = x;
      add[v] = x;
      return;
    }
    push(v, tl, tr);
    int tm = (tl + tr) >> 1;
    upd(v + v, tl, tm, l, min(r, tm), x);
    upd(v + v + 1, tm + 1, tr, max(l, tm + 1), r, x);
    t[v] = max(t[v + v], t[v + v + 1]);
  }
  int get(int v, int tl, int tr, int l, int r) {
    if (l > r) return 0;
    if (tl == l && tr == r) return t[v];
    push(v, tl, tr);
    int tm = (tl + tr) >> 1;
    return max(get(v + v, tl, tm, l, min(tm, r)),
               get(v + v + 1, tm + 1, tr, max(tm + 1, l), r));
  }
  void solve() {
    cin >> n >> m;
    map<int, int> com;
    for (int i = 0; i < m; i++) {
      int id, l, r;
      cin >> id >> l >> r;
      com[l]++, com[r]++;
      g[id].push_back({l, r});
    }
    int val = 1;
    for (auto &it : com) it.second = val++;
    for (int i = 1; i <= n; i++)
      for (auto &it : g[i]) it.first = com[it.first], it.second = com[it.second];
    int nn = val - 1;
    dp[1] = 1;
    for (auto iv : g[1]) upd(1, 1, nn, iv.first, iv.second, 1);
    for (int i = 2; i <= n; i++) {
      dp[i] = 1;
      for (auto iv : g[i])
        dp[i] = max(dp[i], get(1, 1, nn, iv.first, iv.second) + 1);
      for (auto iv : g[i]) upd(1, 1, nn, iv.first, iv.second, dp[i]);
    }
    int ans = *max_element(dp + 1, dp + n + 1), last = -1;
    cout << n - ans << '\n';
    for (int i = n; i >= 1; i--) {
      bool ok = 0;
      if (last == -1)
        ok = 1;
      else {
        for (auto iv : g[i])
          for (auto lastv : g[last])
            if (max(iv.first, lastv.first) <= min(iv.second, lastv.second))
              ok = 1;
      }
      if (!ok || dp[i] != ans)
        cout << i << ' ';
      else
        last = i, ans--;
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int tt = 1;
    while (tt--) {
      solve();
      cout << '\n';
    }
    return 0;
  }
  