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;
  using pii = pair<int, int>;
  struct SegTree {
    vector<pii> st;
    vector<pii> lazy;
    SegTree(int n) : st(4 * n + 1), lazy(4 * n + 1){};
    void push(int v) {
      for (int ch = v * 2; ch <= v * 2 + 1; ch++) {
        st[ch] = max(st[ch], lazy[v]);
        lazy[ch] = max(lazy[ch], lazy[v]);
      }
      lazy[v] = {0, 0};
    }
    void update(int v, int tl, int tr, int l, int r, pii val) {
      if (l > r) return;
      if (tl == l && tr == r) {
        st[v] = max(val, st[v]);
        lazy[v] = max(val, lazy[v]);
        return;
      }
      push(v);
      int tm = (tl + tr) / 2;
      update(v * 2, tl, tm, l, min(r, tm), val);
      update(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r, val);
      st[v] = max(st[v * 2], st[v * 2 + 1]);
    }
    pii query(int v, int tl, int tr, int l, int r) {
      if (l > r) return {0, 0};
      if (tl == l && tr == r) return st[v];
      push(v);
      int tm = (tl + tr) / 2;
      pii ql = query(v * 2, tl, tm, l, min(r, tm));
      pii qr = query(v * 2 + 1, tm + 1, tr, max(l, tm + 1), r);
      return max(ql, qr);
    }
  };
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    map<int, int> comp;
    vector<vector<pii>> seg(n + 1);
    for (int j = 0; j < m; j++) {
      int i, l, r;
      cin >> i >> l >> r;
      seg[i].push_back({l, r});
      comp[l] = 0;
      comp[r] = 0;
    }
    int num = 0;
    for (auto &[k, v] : comp) {
      v = ++num;
    }
    SegTree st(num);
    vector<int> to(n + 5);
    for (int i = 1; i <= n; i++) {
      pii mx = {0, 0};
      for (auto &[l, r] : seg[i]) {
        l = comp[l], r = comp[r];
        mx = max(mx, st.query(1, 1, num, l, r));
      }
      to[i] = mx.second;
      mx = {mx.first + 1, i};
      for (auto [l, r] : seg[i]) {
        st.update(1, 1, num, l, r, mx);
      }
    }
    auto [ans, r] = st.st[1];
    to[n + 1] = r;
    cout << n - ans << '\n';
    int rw = n + 1;
    while (rw) {
      for (int i = to[rw] + 1; i < rw; i++) cout << i << ' ';
      rw = to[rw];
    }
    cout << '\n';
    return 0;
  }
  