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 N = 3e5 + 5;
  pair<int, int> seg[8 * N], lazy[8 * N];
  void build(int l, int u, int i) {
    lazy[i] = {3 * N, -1};
    seg[i] = {0, -1};
    if (l == u) return;
    build(l, ((l + u) / 2), (i * 2 + 1));
    build(((l + u) / 2) + 1, u, (i * 2 + 2));
  }
  void lazup(int l, int u, int i) {
    if (lazy[i].first == 3 * N) return;
    seg[i] = min(seg[i], lazy[i]);
    if (l != u) {
      lazy[(i * 2 + 1)] = min(lazy[(i * 2 + 1)], lazy[i]);
      lazy[(i * 2 + 2)] = min(lazy[(i * 2 + 2)], lazy[i]);
    }
    lazy[i] = {3 * N, -1};
  }
  pair<int, int> query(int l, int u, int i, int ll, int uu) {
    lazup(l, u, i);
    if (l >= ll && u <= uu) {
      return seg[i];
    }
    if (l > uu || u < ll) return {3 * N, -1};
    return min(query(l, ((l + u) / 2), (i * 2 + 1), ll, uu),
               query(((l + u) / 2) + 1, u, (i * 2 + 2), ll, uu));
  }
  pair<int, int> update(int l, int u, int i, int ll, int uu, pair<int, int> &x) {
    lazup(l, u, i);
    if (l >= ll && u <= uu) {
      lazy[i] = x;
      lazup(l, u, i);
      return seg[i];
    }
    if (l > uu || u < ll) return seg[i];
    return seg[i] = min(update(l, ((l + u) / 2), (i * 2 + 1), ll, uu, x),
                        update(((l + u) / 2) + 1, u, (i * 2 + 2), ll, uu, x));
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int n, m;
    cin >> n >> m;
    vector<pair<int, int>> r[n];
    map<int, int> mp;
    set<int> s;
    for (int i = 0; i < m; i++) {
      int x, ll, rr;
      cin >> x >> ll >> rr;
      r[x - 1].push_back({ll, rr});
      s.insert(ll);
      s.insert(rr);
    }
    int c = 0;
    for (int x : s) {
      mp[x] = c++;
    }
    int oset = 0;
    int dp[n], prev[n];
    build(0, c - 1, 0);
    int ans = n, curr = -1;
    for (int i = 0; i < n; i++) {
      dp[i] = i;
      prev[i] = -1;
      for (pair<int, int> k : r[i]) {
        pair<int, int> lol = query(0, c - 1, 0, mp[k.first], mp[k.second]);
        if ((lol.first + oset) < dp[i]) {
          dp[i] = lol.first + oset;
          prev[i] = lol.second;
        }
      }
      pair<int, int> x = {dp[i] - (oset + 1), i};
      for (pair<int, int> k : r[i]) {
        update(0, c - 1, 0, mp[k.first], mp[k.second], x);
      }
      ans = min(ans, dp[i] + (n - i - 1));
      if (ans == (dp[i] + n - i - 1)) curr = i;
      oset++;
    }
    cout << ans << endl;
    int currind = n - 1;
    while (currind >= 0) {
      while (currind > curr) {
        cout << (currind + 1) << " ";
        currind--;
      }
      if (curr == -1)
        currind = -1;
      else {
        curr = prev[curr];
        currind--;
      }
    }
  }
  