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;
  struct gg {
    int c, fa;
    bool operator<(const gg &QAQ) const { return c < QAQ.c; }
  } g0, ans;
  struct lxy {
    int lson, rson;
    gg v, tg;
  } b[2000005];
  struct lxy1 {
    int x, l, r;
    bool operator<(const lxy1 &QAQ) const { return x < QAQ.x; }
  } q[300005];
  int n, m, cnt, rt, N;
  int g[300005];
  bool vis[300005];
  priority_queue<int> p;
  map<int, int> h;
  void update(int u) { b[u].v = max(b[b[u].lson].v, b[b[u].rson].v); }
  void pushdown(int u) {
    if (b[u].tg.c == 0) return;
    b[b[u].lson].v = b[u].tg, b[b[u].lson].tg = b[u].tg;
    b[b[u].rson].v = b[u].tg, b[b[u].rson].tg = b[u].tg;
    b[u].tg.c = 0;
  }
  void build(int &u, int l, int r) {
    u = ++cnt;
    if (l == r) return;
    int mid = (l + r) / 2;
    build(b[u].lson, l, mid);
    build(b[u].rson, mid + 1, r);
  }
  void modify(int u, int L, int R, int l, int r, int k, int id) {
    if (L == l && R == r) {
      b[u].v.c = k, b[u].tg.c = k;
      b[u].v.fa = id, b[u].tg.fa = id;
      return;
    }
    int mid = (L + R) / 2;
    if (mid >= r)
      pushdown(u), modify(b[u].lson, L, mid, l, r, k, id);
    else if (l > mid)
      pushdown(u), modify(b[u].rson, mid + 1, R, l, r, k, id);
    else
      pushdown(u), modify(b[u].lson, L, mid, l, mid, k, id),
          modify(b[u].rson, mid + 1, R, mid + 1, r, k, id);
    update(u);
  }
  gg query(int u, int L, int R, int l, int r) {
    pushdown(u);
    if (L == l && R == r) return b[u].v;
    int mid = (L + R) / 2;
    if (mid >= r) {
      return query(b[u].lson, L, mid, l, r);
    } else if (l > mid) {
      return query(b[u].rson, mid + 1, R, l, r);
    } else {
      return max(query(b[u].lson, L, mid, l, mid),
                 query(b[u].rson, mid + 1, R, mid + 1, r));
    }
    update(u);
  }
  int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; i++) {
      scanf("%d%d%d", &q[i].x, &q[i].l, &q[i].r);
      p.push(-q[i].l), p.push(-q[i].r);
    }
    sort(q + 1, q + 1 + m);
    N = p.size();
    build(rt, 1, p.size());
    cnt = 0;
    while (!p.empty()) {
      int y = p.top();
      h[-p.top()] = ++cnt;
      p.pop();
      while (!p.empty() && p.top() == y) p.pop();
    }
    for (int i = 1; i <= m; i++) {
      gg ret = g0;
      int now = i;
      while (q[now].x == q[now + 1].x) {
        ret = max(ret, query(rt, 1, N, h[q[now].l], h[q[now].r]));
        now++;
      }
      ret = max(ret, query(rt, 1, N, h[q[now].l], h[q[now].r]));
      for (int j = i; j <= now; j++)
        modify(rt, 1, N, h[q[j].l], h[q[j].r], ret.c + 1, q[j].x);
      i = now;
      g[q[i].x] = ret.fa;
      if (ret.c + 1 > ans.c) ans.c = ret.c + 1, ans.fa = q[i].x;
    }
    int pos = ans.fa;
    while (pos != 0) {
      vis[pos] = 1;
      pos = g[pos];
    }
    cout << n - ans.c << endl;
    for (int i = 1; i <= n; i++)
      if (vis[i] == 0) printf("%d ", i);
  }
  