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;
  struct segtree {
    int size = 1;
    vector<pair<int, int>> v;
    vector<pair<int, int>> lazy;
    void init(int n) {
      while (n > size) {
        size *= 2;
      }
      v.assign(size * 2, {0, -1});
      lazy.assign(size * 2, {0, -1});
    }
    void prop(int x, int lx, int rx) {
      if (rx - lx == 1) return;
      if (lazy[x] == make_pair(0, -1)) return;
      v[x * 2 + 1] = lazy[x];
      v[x * 2 + 2] = lazy[x];
      lazy[x * 2 + 1] = lazy[x];
      lazy[x * 2 + 2] = lazy[x];
      lazy[x] = {0, -1};
    }
    void update(int l, int r, pair<int, int> val, int x, int lx, int rx) {
      prop(x, lx, rx);
      if (rx <= l || lx >= r) {
        return;
      }
      if (lx >= l && rx <= r) {
        v[x] = val;
        lazy[x] = val;
        return;
      }
      int m = (lx + rx) / 2;
      update(l, r, val, x * 2 + 1, lx, m);
      update(l, r, val, x * 2 + 2, m, rx);
      v[x] = max(v[x * 2 + 1], v[x * 2 + 2]);
    }
    void update(int l, int r, pair<int, int> val) {
      update(l, r + 1, val, 0, 0, size);
    }
    pair<int, int> get(int l, int r, int x, int lx, int rx) {
      prop(x, lx, rx);
      if (rx <= l || lx >= r) {
        return {0, -1};
      }
      if (lx >= l && rx <= r) {
        return v[x];
      }
      int m = (lx + rx) / 2;
      pair<int, int> m1 = get(l, r, x * 2 + 1, lx, m);
      pair<int, int> m2 = get(l, r, x * 2 + 2, m, rx);
      return max(m1, m2);
    }
    pair<int, int> get(int l, int r) { return get(l, r + 1, 0, 0, size); }
  };
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n, m;
    cin >> n >> m;
    set<int> s;
    vector<vector<pair<int, int>>> v(n);
    vector<tuple<int, int, int>> actions;
    for (int i = 0; i < m; i++) {
      int j, l, r;
      cin >> j >> l >> r;
      j--;
      actions.push_back({j, l, r});
      s.insert(l);
      s.insert(r);
      v[j].push_back(make_pair(l, r));
    }
    auto s1 = s.begin();
    auto s2 = s.begin();
    if (s2 != s.end()) s2++;
    map<int, int> mapa;
    int ctr = 0;
    mapa[*s1] = ctr;
    ctr++;
    while (s2 != s.end()) {
      if ((*s1) != (*s2)) {
        mapa[*s2] = ctr;
        ctr++;
      }
      s1++;
      s2++;
    }
    segtree prev;
    prev.init(2 * m + 2);
    vector<int> before(n, -1);
    for (int i = 0; i < n; i++) {
      pair<int, int> mx = {0, -1};
      for (auto x : v[i]) {
        int a = mapa[x.first];
        int b = mapa[x.second];
        mx = max(mx, prev.get(a, b));
      }
      mx.first++;
      before[i] = mx.second;
      mx.second = i;
      for (auto x : v[i]) {
        int a = mapa[x.first];
        int b = mapa[x.second];
        prev.update(a, b, mx);
      }
    }
    vector<int> ok(n, false);
    int lastmx = prev.get(0, 2 * m + 1).second;
    while (lastmx >= 0) {
      ok[lastmx] = true;
      lastmx = before[lastmx];
    }
    vector<int> res;
    int resnum = 0;
    for (int i = 0; i < n; i++) {
      if (!ok[i]) {
        res.push_back(i);
        resnum++;
      }
    }
    cout << resnum << '\n';
    for (auto w : res) cout << w + 1 << " ";
  }
  