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 int N = 3e5 + 5;
  int n, m, opt, l, r, v, ans;
  int minn[N * 40], tag[N * 40], lson[N * 40], rson[N * 40], cnt = 1;
  inline void pushdown(int k) {
    if (~tag[k]) {
      minn[lson[k]] = minn[rson[k]] = tag[k];
      tag[lson[k]] = tag[rson[k]] = tag[k];
      tag[k] = -1;
    }
  }
  void change(int k, int l, int r, int qx, int qy, int v) {
    if (qx <= l && r <= qy) {
      minn[k] = v;
      tag[k] = v;
      return;
    }
    if (!lson[k]) lson[k] = ++cnt;
    if (!rson[k]) rson[k] = ++cnt;
    pushdown(k);
    int mid = l + r >> 1;
    if (qx <= mid) change(lson[k], l, mid, qx, qy, v);
    if (mid < qy) change(rson[k], mid + 1, r, qx, qy, v);
    minn[k] = max(minn[lson[k]], minn[rson[k]]);
  }
  int query(int k, int l, int r, int qx, int qy) {
    if (qx <= l && r <= qy) return minn[k];
    if (!lson[k]) lson[k] = ++cnt;
    if (!rson[k]) rson[k] = ++cnt;
    pushdown(k);
    int mid = l + r >> 1;
    int ans = 0;
    if (qx <= mid) ans = max(ans, query(lson[k], l, mid, qx, qy));
    if (mid < qy) ans = max(ans, query(rson[k], mid + 1, r, qx, qy));
    return ans;
  }
  vector<pair<int, int>> q[300500];
  map<int, int> mp;
  int nxt[300500], i, j, f[300500], mx, res[300500];
  void work(int x) {
    res[x] = 1;
    int i, j, k;
    change(1, 1, 1000000000, 1, 1000000000, 0);
    for (auto [i, j] : q[x]) {
      change(1, 1, 1000000000, i, j, 1);
    }
    for (j = x + 1; j <= n; j++) {
      if (f[j] != f[x] - 1) {
        continue;
      }
      k = 0;
      for (auto [l1, r1] : q[j]) {
        k = max(k, query(1, 1, 1000000000, l1, r1));
      }
      if (k) {
        work(j);
        return;
      }
    }
  }
  int main() {
    memset(tag, -1, sizeof(tag));
    scanf("%d%d", &n, &m);
    for (i = 1; i <= m; i++) {
      int x, y, z;
      scanf("%d%d%d", &x, &y, &z);
      q[x].push_back({y, z});
    }
    for (i = n; i >= 1; i--) {
      int sb = 0;
      for (auto [x, y] : q[i]) {
        sb = max(sb, query(1, 1, 1000000000, x, y));
      }
      nxt[i] = mp[sb];
      sb++;
      for (auto [x, y] : q[i]) {
        change(1, 1, 1000000000, x, y, sb);
      }
      mp[sb] = i;
      f[i] = sb;
      mx = max(mx, sb);
    }
    for (i = 1; i <= n; i++) {
      if (f[i] == mx) {
        work(i);
        break;
      }
    }
    printf("%d\n", n - mx);
    for (i = 1; i <= n; i++) {
      if (!res[i]) {
        printf("%d ", i);
      }
    }
  }
  