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 std::vector;
  struct op {
    int val;
    int pos;
    op() { val = pos = 0; }
  };
  op max(op u, op v) {
    if (u.val > v.val)
      return u;
    else
      return v;
  }
  struct SegmentT {
    int l, r;
    op d;
    op c;
  };
  SegmentT t[4000005];
  struct Element {
    int l, r;
  };
  int a[1000005];
  vector<Element> g[300005];
  int nxt[300005];
  int vis[300005];
  int preWork(int n) {
    std::sort(a + 1, a + 1 + n);
    int p = 1, size = 0;
    while (p <= n) {
      int np = p;
      while (np + 1 <= n && a[np + 1] == a[p]) np++;
      a[++size] = a[p];
      p = np + 1;
    }
    return size;
  }
  int getID(int u, int n) {
    int L = 1, R = n;
    int res = -1;
    while (L <= R) {
      int M = (L + R) / 2;
      if (a[M] <= u) {
        res = M;
        L = M + 1;
      } else
        R = M - 1;
    }
    return res;
  }
  int max(int a, int b) { return a > b ? a : b; }
  void build(int p, int l, int r) {
    t[p].l = l, t[p].r = r;
    t[p].c.val = t[p].c.pos = 0;
    if (l == r)
      t[p].d.val = t[p].d.pos = 0;
    else {
      int m = (l + r) / 2;
      build(2 * p, l, m);
      build(2 * p + 1, m + 1, r);
      t[p].d = max(t[2 * p].d, t[2 * p + 1].d);
    }
  }
  void spread(int p) {
    if (t[p].c.val != 0) {
      op c = t[p].c;
      t[p].c.val = t[p].c.pos = 0;
      t[2 * p].d = max(t[2 * p].d, c);
      t[2 * p].c = max(t[2 * p].c, c);
      t[2 * p + 1].d = max(t[2 * p + 1].d, c);
      t[2 * p + 1].c = max(t[2 * p + 1].c, c);
    }
  }
  void change(int p, int l, int r, op v) {
    if (l <= t[p].l && r >= t[p].r) {
      t[p].d = max(t[p].d, v);
      t[p].c = max(t[p].c, v);
    } else {
      spread(p);
      int m = (t[p].l + t[p].r) / 2;
      if (l <= m) change(2 * p, l, r, v);
      if (r >= m + 1) change(2 * p + 1, l, r, v);
      t[p].d = max(t[2 * p].d, t[2 * p + 1].d);
    }
  }
  op ask(int p, int l, int r) {
    if (l <= t[p].l && r >= t[p].r)
      return t[p].d;
    else {
      spread(p);
      op res;
      res.val = res.pos = 0;
      int m = (t[p].l + t[p].r) / 2;
      if (l <= m) res = max(res, ask(2 * p, l, r));
      if (r >= m + 1) res = max(res, ask(2 * p + 1, l, r));
      return res;
    }
  }
  int main() {
    int n, m;
    scanf("%d%d", &n, &m);
    int size = 0;
    for (int i = 0; i < m; i++) {
      int r, x, y;
      scanf("%d%d%d", &r, &x, &y);
      a[++size] = x;
      a[++size] = y + 1;
      g[r].push_back(Element{x, y});
    }
    size = preWork(size);
    build(1, 1, size);
    int ans = 1, id = 1;
    for (int i = n; i >= 1; i--) {
      int val = 1;
      nxt[i] = 0;
      for (Element e : g[i]) {
        int l = e.l, r = e.r;
        int x = getID(l, size);
        int y = getID(r, size);
        op offer = ask(1, x, y);
        if (val < offer.val + 1) {
          val = offer.val + 1;
          nxt[i] = offer.pos;
        }
      }
      if (ans < val) {
        ans = val;
        id = i;
      }
      for (Element e : g[i]) {
        int l = e.l, r = e.r;
        int x = getID(l, size);
        int y = getID(r, size);
        op now;
        now.val = val, now.pos = i;
        change(1, x, y, now);
      }
    }
    int p = id;
    while (p != 0) {
      vis[p] = 1;
      p = nxt[p];
    }
    printf("%d\n", n - ans);
    for (int i = 1; i <= n; i++) {
      if (!vis[i]) printf("%d ", i);
    }
    printf("\n");
    return 0;
  }
  