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 MN = 300005;
  long long n, m, row, l, r, dp[MN], bck[MN];
  vector<pair<long long, long long> > v[MN];
  vector<long long> xs;
  map<long long, long long> mp;
  pair<long long, long long> st[MN * 12], lzy[MN * 12];
  pair<long long, long long> push_up(pair<long long, long long> x,
                                     pair<long long, long long> y) {
    if (x.first < y.first)
      return x;
    else
      return y;
  }
  void push_down(long long l, long long r, long long idx) {
    if (!lzy[idx].second) return;
    st[idx] = min(st[idx], lzy[idx]);
    if (l != r) lzy[idx * 2] = lzy[idx * 2 + 1] = min(lzy[idx * 2], lzy[idx]);
    lzy[idx] = {0, 0};
  }
  void build(long long l, long long r, long long idx) {
    if (l == r) {
      st[idx] = {-1, 0};
      return;
    }
    long long mid = (l + r) >> 1;
    build(l, mid, idx * 2), build(mid + 1, r, idx * 2 + 1);
    st[idx] = push_up(st[idx * 2], st[idx * 2 + 1]);
  }
  void upd(long long l, long long r, long long x, long long y,
           pair<long long, long long> val, long long idx) {
    push_down(l, r, idx);
    if (r < x || l > y) return;
    if (r <= y && l >= x) {
      lzy[idx] = val;
      push_down(l, r, idx);
      return;
    }
    long long mid = (l + r) >> 1;
    upd(l, mid, x, y, val, idx * 2), upd(mid + 1, r, x, y, val, idx * 2 + 1);
    st[idx] = push_up(st[idx * 2], st[idx * 2 + 1]);
  }
  pair<long long, long long> qry(long long l, long long r, long long x,
                                 long long y, long long idx) {
    push_down(l, r, idx);
    if (r < x || l > y) return {0x3f3f3f3f, 0x3f3f3f3f};
    if (r <= y && l >= x) return st[idx];
    long long mid = (l + r) >> 1;
    return push_up(qry(l, mid, x, y, idx * 2),
                   qry(mid + 1, r, x, y, idx * 2 + 1));
  }
  int32_t main() {
    cin.tie(0), cin.sync_with_stdio(0);
    cin >> n >> m;
    for (long long i = 1; i <= m; i++) {
      cin >> row >> l >> r;
      xs.push_back(l);
      xs.push_back(r);
      v[row].push_back({l, r});
    }
    sort(xs.begin(), xs.end());
    xs.erase(unique(xs.begin(), xs.end()), xs.end());
    for (long long i = 0; i < xs.size(); i++) mp[xs[i]] = i + 1;
    v[n + 1].push_back({xs[0], xs.back()});
    build(1, xs.size(), 1);
    for (long long i = 1; i <= n + 1; i++) {
      pair<long long, long long> val = {-1, 0};
      for (pair<long long, long long> p : v[i]) {
        l = mp[p.first], r = mp[p.second];
        pair<long long, long long> tmp = qry(1, xs.size(), l, r, 1);
        val = min(val, tmp);
      }
      dp[i] = val.first + i;
      bck[i] = val.second;
      for (pair<long long, long long> p : v[i]) {
        l = mp[p.first], r = mp[p.second];
        upd(1, xs.size(), l, r, {dp[i] - i - 1, i}, 1);
      }
    }
    printf("%lld\n", dp[n + 1]);
    vector<long long> out;
    long long cur = n + 1;
    while (cur) {
      long long pre = bck[cur];
      for (long long i = pre + 1; i < cur; i++) out.push_back(i);
      cur = pre;
    }
    sort(out.begin(), out.end());
    for (long long p : out) printf("%lld ", p);
    return 0;
  }
  