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;
  int n, m, cnt, A[600009 << 1], f[600009], idx[600009];
  struct data {
    int x, l, r;
  } a[600009];
  vector<data> cc[600009];
  struct node {
    int id, mi;
    node() { id = 0, mi = 1000000000; }
  } s[600009 << 2], tag[600009 << 2];
  node operator+(node i, node j) {
    node o;
    if (i.mi < j.mi) {
      o = i;
    } else {
      o = j;
    }
    return o;
  }
  void mark(int now, node v) {
    if (v.mi < s[now].mi) {
      s[now] = v;
      tag[now] = v;
    } else {
      tag[now] = v;
    }
  }
  void pushdown(int now) {
    if (tag[now].mi != 1000000000) {
      mark((now << 1), tag[now]);
      mark((now << 1 | 1), tag[now]);
      tag[now] = node();
    }
  }
  void update(int l, int r, int now, int L, int R, node v) {
    if (l >= L && r <= R) {
      mark(now, v);
      return;
    }
    pushdown(now);
    int mid = (l + r) >> 1;
    if (L <= mid) update(l, mid, (now << 1), L, R, v);
    if (R > mid) update(mid + 1, r, (now << 1 | 1), L, R, v);
    s[now] = s[(now << 1)] + s[(now << 1 | 1)];
  }
  node query(int l, int r, int now, int L, int R) {
    if (l >= L && r <= R) {
      return s[now];
    }
    pushdown(now);
    int mid = (l + r) >> 1;
    if (L <= mid && R > mid)
      return query(l, mid, (now << 1), L, R) +
             query(mid + 1, r, (now << 1 | 1), L, R);
    else if (L <= mid)
      return query(l, mid, (now << 1), L, R);
    else
      return query(mid + 1, r, (now << 1 | 1), L, R);
  }
  int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; ++i) {
      scanf("%d%d%d", &a[i].x, &a[i].l, &a[i].r);
      A[++cnt] = a[i].l, A[++cnt] = a[i].r;
    }
    sort(A + 1, A + 1 + cnt);
    cnt = unique(A + 1, A + 1 + cnt) - A - 1;
    for (int i = 1; i <= m; ++i) {
      a[i].l = lower_bound(A + 1, A + 1 + cnt, a[i].l) - A;
      a[i].r = lower_bound(A + 1, A + 1 + cnt, a[i].r) - A;
      cc[a[i].x].push_back(a[i]);
    }
    int ans = n - 1, ii = 0;
    for (int i = n; i >= 1; --i) {
      f[i] = n - i;
      int id = n + 1;
      idx[i] = n + 1;
      if (!cc[i].size()) {
        continue;
      }
      for (int j = 0; j < cc[i].size(); ++j) {
        int l = cc[i][j].l, r = cc[i][j].r;
        node pp = query(1, cnt, 1, l, r);
        if (pp.mi - i - 1 < f[i]) {
          f[i] = pp.mi - i - 1;
          id = pp.id;
        }
      }
      idx[i] = id;
      node nn;
      nn.mi = f[i] + i, nn.id = i;
      for (int j = 0; j < cc[i].size(); ++j) {
        int l = cc[i][j].l, r = cc[i][j].r;
        update(1, cnt, 1, l, r, nn);
      }
      if (i - 1 + f[i] < ans) ii = i;
      ans = min(ans, i - 1 + f[i]);
    }
    if (ans == n - 1) {
      printf("%d\n", ans);
      for (int i = 1; i < n; ++i) printf("%d ", i);
      return 0;
    }
    if (ans == 0)
      printf("0\n");
    else {
      printf("%d\n", ans);
      for (int i = 1; i < ii; ++i) printf("%d ", i);
      for (int i = ii; i <= n; i = idx[i]) {
        int l = i + 1, r = idx[i] - 1;
        for (int p = l; p <= r; ++p) printf("%d ", p);
      }
    }
    return 0;
  }
  