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 int mod = 1000000007;
  const int mxx = 300005;
  vector<pair<int, int> > v[2 * mxx];
  unordered_map<int, int> mp;
  long long int Max[mxx];
  long long int tree[8 * mxx][2];
  void update(int l, int r, int L, int R, int n, long long int val) {
    if (l > R || r < L)
      return;
    else if (l >= L and r <= R) {
      tree[n][0] = val + 1ll;
      tree[n][1] = val + 1;
      return;
    }
    update(l, (l + r) / 2, L, R, n * 2, val);
    update((l + r) / 2 + 1, r, L, R, n * 2 + 1, val);
    tree[n][0] = max(tree[n * 2][0], tree[n * 2 + 1][0]);
  }
  long long int go(int l, int r, int L, int R, int n, long long int val) {
    if (l > R || r < L)
      return 0;
    else if (l >= L and r <= R) {
      return tree[n][0];
    }
    val = max(val, tree[n][1]);
    long long int x = max(go(l, (l + r) / 2, L, R, n * 2, val),
                          go((l + r) / 2 + 1, r, L, R, n * 2 + 1, val));
    return max(val, x);
  }
  void solve() {
    int n, m;
    cin >> n >> m;
    vector<int> compress, del;
    for (int i = 1; i <= m; i++) {
      int a, b, c;
      cin >> a >> b >> c;
      v[a].push_back({b, c});
      compress.push_back(b);
      compress.push_back(c);
    }
    sort(compress.begin(), compress.end());
    int sum = 0;
    for (auto it : compress) {
      if (mp[it] == 0) mp[it] = ++sum;
    }
    long long int ans = 0;
    int p = 0;
    for (int i = n; i >= 1; i--) {
      long long int mx = 0;
      for (auto it : v[i]) {
        int x = mp[it.first];
        int y = mp[it.second];
        long long int fre = go(1, sum, x, y, 1, 0);
        mx = max(mx, fre);
      }
      for (auto it : v[i]) {
        int x = mp[it.first];
        int y = mp[it.second];
        update(1, sum, x, y, 1, mx);
      }
      Max[i] = mx + 1;
      if (ans < mx + 1) {
        ans = mx + 1;
        p = i;
      }
    }
    memset(tree, 0, sizeof(tree));
    for (int i = 1; i <= n; i++) {
      if (i < p) {
        del.push_back(i);
      } else if (i == p) {
        for (auto it : v[i]) {
          update(1, sum, mp[it.first], mp[it.second], 1, 1);
        }
      } else if (i > p) {
        if (Max[i] < ans - 1) {
          del.push_back(i);
          continue;
        }
        int mx = 0;
        for (auto it : v[i]) {
          int x = mp[it.first];
          int y = mp[it.second];
          int fre = go(1, sum, x, y, 1, 0);
          mx = max(mx, fre);
        }
        if (mx) {
          for (auto it : v[p]) {
            update(1, sum, mp[it.first], mp[it.second], 1, -1);
          }
          ans--;
          for (auto it : v[i]) {
            update(1, sum, mp[it.first], mp[it.second], 1, 1);
          }
          p = i;
        } else
          del.push_back(i);
      }
    }
    cout << del.size() << "\n";
    for (auto it : del) {
      cout << it << " ";
    }
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(NULL);
    int t = 1;
    while (t--) solve();
  }
  