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 maxn = 300005;
  int n, m, need[maxn];
  vector<pair<int, int> > row[maxn];
  int dp[maxn], tr[maxn];
  pair<int, int> tree[maxn * 8];
  pair<int, int> lazy[maxn * 8];
  void PushUp(int node) {
    if (tree[node * 2] > tree[node * 2 + 1])
      tree[node] = tree[node * 2];
    else
      tree[node] = tree[node * 2 + 1];
  }
  void PushDown(int node, int l, int r) {
    if (tree[node].first < lazy[node].first) tree[node] = lazy[node];
    if (l == r) return;
    if (lazy[node * 2].first < lazy[node].first) {
      lazy[node * 2] = lazy[node];
    }
    if (lazy[node * 2 + 1].first < lazy[node].first) {
      lazy[node * 2 + 1] = lazy[node];
    }
    lazy[node] = {0, 0};
  }
  void Update(int node, int l, int r, int u, int v, int id, int val) {
    PushDown(node, l, r);
    if (l > v || r < u) return;
    if (l >= u && r <= v) {
      if (tree[node].first < val) {
        tree[node] = {val, id};
      }
      if (lazy[node].first < val) {
        lazy[node] = {val, id};
      }
      return;
    }
    int mid = (l + r) >> 1;
    Update(node * 2, l, mid, u, v, id, val);
    Update(node * 2 + 1, mid + 1, r, u, v, id, val);
    PushUp(node);
  }
  pair<int, int> Get(int node, int l, int r, int u, int v) {
    PushDown(node, l, r);
    if (l > v || r < u) return {0, 0};
    if (l >= u && r <= v) return tree[node];
    int mid = (l + r) >> 1;
    pair<int, int> L = Get(node * 2, l, mid, u, v);
    pair<int, int> R = Get(node * 2 + 1, mid + 1, r, u, v);
    if (L.first > R.first) return L;
    return R;
  }
  void Solve(int Test) {
    cin >> n >> m;
    vector<int> v;
    for (int i = 1; i <= m; ++i) {
      int id, l, r;
      cin >> id >> l >> r;
      v.push_back(l), v.push_back(r);
      row[id].push_back({l, r});
    }
    sort(v.begin(), v.end());
    v.resize(unique(v.begin(), v.end()) - v.begin());
    m = v.size();
    for (int i = 1; i <= n; ++i) {
      for (auto &p : row[i]) {
        int &l = p.first, &r = p.second;
        l = lower_bound(v.begin(), v.end(), l) - v.begin() + 1;
        r = lower_bound(v.begin(), v.end(), r) - v.begin() + 1;
        pair<int, int> cur = Get(1, 1, m, l, r);
        if (dp[i] < cur.first + 1) {
          dp[i] = cur.first + 1;
          tr[i] = cur.second;
        }
      }
      for (auto p : row[i]) {
        int l = p.first, r = p.second;
        Update(1, 1, m, l, r, i, dp[i]);
      }
    }
    int pos = 0;
    for (int i = 1; i <= n; ++i) {
      if (dp[i] > dp[pos]) {
        pos = i;
      }
    }
    int res = 0;
    while (pos) {
      need[pos] = 1;
      pos = tr[pos];
      res++;
    }
    cout << n - res << "\n";
    for (int i = 1; i <= n; ++i) {
      if (!need[i]) cout << i << " ";
    }
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    if (fopen("CF"
              ".inp",
              "r")) {
      freopen(
          "CF"
          ".inp",
          "r", stdin);
      freopen(
          "CF"
          ".out",
          "w", stdout);
    }
    int test = 1;
    for (int i = 1; i <= test; ++i) {
      Solve(i);
    }
  }
  