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 inf = 1e9;
  const long long mod = 1e9 + 7;
  struct node {
    long long id, l, r;
  } a[300005];
  struct node1 {
    pair<long long, long long> val, lazy;
  } tree[2400005];
  long long b[600005], pre[300005];
  bool vis[300005];
  vector<pair<long long, long long> > g[300005];
  bool cmp(node a, node b) { return a.id < b.id; }
  void up(long long p) {
    tree[p].val = max(tree[2 * p].val, tree[2 * p + 1].val);
  }
  void pushdown(long long p) {
    if (tree[p].lazy.first != 0) {
      tree[2 * p].lazy = tree[2 * p + 1].lazy = tree[p].lazy;
      tree[2 * p].val = tree[2 * p + 1].val = tree[p].lazy;
      tree[p].lazy = {0, 0};
    }
  }
  void update(long long l, long long r, long long p, pair<long long, long long> d,
              long long cl, long long cr) {
    if (cl > r || cr < l)
      return;
    else if (cl >= l && cr <= r) {
      tree[p].val = d;
      tree[p].lazy = d;
    } else {
      pushdown(p);
      long long mid = (cl + cr) / 2;
      update(l, r, 2 * p, d, cl, mid);
      update(l, r, 2 * p + 1, d, mid + 1, cr);
      up(p);
    }
  }
  pair<long long, long long> query(long long l, long long r, long long p,
                                   long long cl, long long cr) {
    if (cl > r || cr < l)
      return {0, 0};
    else if (cl >= l && cr <= r) {
      return tree[p].val;
    } else {
      pushdown(p);
      long long mid = (cl + cr) / 2;
      return max(query(l, r, 2 * p, cl, mid),
                 query(l, r, 2 * p + 1, mid + 1, cr));
    }
  }
  void solve() {
    long long n, m, cnt = 0, ans = 0, flag = 0;
    cin >> n >> m;
    for (long long i = 1; i <= m; i++) {
      cin >> a[i].id >> a[i].l >> a[i].r;
      b[++cnt] = a[i].l, b[++cnt] = a[i].r;
    }
    sort(b + 1, b + 1 + cnt);
    cnt = unique(b + 1, b + 1 + cnt) - b - 1;
    for (long long i = 1; i <= m; i++) {
      a[i].l = lower_bound(b + 1, b + 1 + cnt, a[i].l) - b;
      a[i].r = lower_bound(b + 1, b + 1 + cnt, a[i].r) - b;
      g[a[i].id].push_back(make_pair(a[i].l, a[i].r));
    }
    for (long long i = 1; i <= n; i++) {
      long long ma = -1;
      for (auto c : g[i]) {
        pair<long long, long long> t = query(c.first, c.second, 1, 1, cnt);
        if (t.first > ma) {
          ma = t.first;
          pre[i] = t.second;
        }
      }
      if (ma + 1 > ans) {
        ans = ma + 1;
        flag = i;
      }
      for (auto c : g[i]) {
        update(c.first, c.second, 1, {ma + 1, i}, 1, cnt);
      }
    }
    cout << n - ans << '\n';
    while (flag != 0) {
      vis[flag] = 1;
      flag = pre[flag];
    }
    for (long long i = 1; i <= n; i++) {
      if (vis[i] == 0) cout << i << ' ';
    }
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  