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 long long MOD = 1e9 + 7;
  const long long N = 6e5 + 10;
  int n;
  vector<pair<int, int>> t(4 * N), lazy(4 * N);
  void push(int v) {
    if (t[v * 2].first < lazy[v].first) {
      t[v * 2] = lazy[v];
    }
    if (lazy[v * 2].first < lazy[v].first) {
      lazy[v * 2] = lazy[v];
    }
    if (t[v * 2 + 1].first < lazy[v].first) {
      t[v * 2 + 1] = lazy[v];
    }
    if (lazy[v * 2 + 1].first < lazy[v].first) {
      lazy[v * 2 + 1] = lazy[v];
    }
    lazy[v] = {0, 0};
  }
  pair<int, int> maxx(int l, int r, int v = 1, int tl = 0, int tr = n - 1) {
    if (l > r) return {0, 0};
    if (l == tl && r == tr) return t[v];
    push(v);
    int tm = (tl + tr) >> 1;
    pair<int, int> lf = maxx(l, min(r, tm), v * 2, tl, tm);
    pair<int, int> rt = maxx(max(l, tm + 1), r, v * 2 + 1, tm + 1, tr);
    if (lf.first > rt.first) {
      return lf;
    } else {
      return rt;
    }
  }
  void update(int l, int r, pair<int, int> val, int v = 1, int tl = 0,
              int tr = n - 1) {
    if (l > r) return;
    if (l == tl && r == tr) {
      if (t[v].first < val.first) {
        t[v] = val;
      }
      if (lazy[v].first < val.first) {
        lazy[v] = val;
      }
    } else {
      push(v);
      int tm = (tl + tr) >> 1;
      update(l, min(r, tm), val, v * 2, tl, tm);
      update(max(l, tm + 1), r, val, v * 2 + 1, tm + 1, tr);
      if (t[v * 2].first > t[v * 2 + 1].first) {
        t[v] = t[v * 2];
      } else {
        t[v] = t[v * 2 + 1];
      }
    }
  }
  void test_case() {
    int nn, m;
    cin >> nn >> m;
    vector<vector<pair<int, int>>> mm(nn + 1);
    map<int, int> zz, id;
    for (int i = 0; i < m; i++) {
      int x, y, z;
      cin >> x >> y >> z;
      mm[x].push_back({y, z});
      zz[y]++;
      zz[z]++;
    }
    n = zz.size();
    int curr = 0;
    for (auto u : zz) {
      id[u.first] = curr++;
    }
    vector<int> par(nn + 1, 0);
    for (int i = 1; i <= nn; i++) {
      for (auto u : mm[i]) {
        update(id[u.first], id[u.second], pair<int, int>{1, i});
      }
    }
    for (int i = 1; i <= nn; i++) {
      int mx = 0;
      for (auto u : mm[i]) {
        pair<int, int> q = maxx(id[u.first], id[u.second]);
        if (q.first >= mx) {
          mx = q.first;
          par[i] = q.second;
        }
      }
      if (mx != 0)
        for (auto u : mm[i]) {
          update(id[u.first], id[u.second], pair<int, int>{mx + 1, i});
        }
    }
    pair<int, int> ans = maxx(0, n - 1);
    curr = ans.second;
    vector<int> done(nn + 1, 0);
    while (curr != 0) {
      done[curr] = 1;
      if (curr == par[curr]) {
        break;
      }
      curr = par[curr];
    }
    vector<int> res;
    for (int i = 1; i <= nn; i++) {
      if (!done[i]) {
        res.push_back(i);
      }
    }
    cout << res.size() << '\n';
    for (auto u : res) {
      cout << u << ' ';
    }
    cout << '\n';
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int tc = 1;
    while (tc--) test_case();
    return 0;
  }
  