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 = 6e5 + 1067, md = 1e9 + 7;
  vector<int> val, to[N];
  map<int, int> mapVal;
  int l[N], r[N], trai[N * 4], phai[N * 4], trace[N];
  int n, m;
  pair<int, int> it[N * 4], lz[N * 4];
  pair<int, int> mp(int x, int y) { return make_pair(x, y); }
  void minimize(pair<int, int> &x, pair<int, int> y) {
    if (x.first > y.first) {
      x.first = y.first;
      x.second = y.second;
    }
  }
  void build(int x, int l, int r) {
    trai[x] = l;
    phai[x] = r;
    it[x] = mp(-1, 0);
    if (l == r) return;
    int mid = (l + r) / 2;
    build(x * 2, l, mid);
    build(x * 2 + 1, mid + 1, r);
  }
  void down(int x) {
    int a = x * 2;
    minimize(it[a], lz[x]);
    minimize(lz[a], lz[x]);
    a += 1;
    minimize(it[a], lz[x]);
    minimize(lz[a], lz[x]);
  }
  void update(int x, int l, int r, pair<int, int> c) {
    if (trai[x] > r || phai[x] < l) return;
    if (trai[x] >= l && phai[x] <= r) {
      minimize(it[x], c);
      minimize(lz[x], c);
      return;
    }
    down(x);
    update(x * 2, l, r, c);
    update(x * 2 + 1, l, r, c);
    minimize(it[x], it[x * 2]);
    minimize(it[x], it[x * 2 + 1]);
  }
  pair<int, int> query(int x, int l, int r) {
    if (trai[x] > r || phai[x] < l) return mp(-1, 0);
    if (trai[x] >= l && phai[x] <= r) {
      return it[x];
    }
    pair<int, int> ans = mp(0, -1);
    down(x);
    minimize(ans, query(x * 2, l, r));
    minimize(ans, query(x * 2 + 1, l, r));
    return ans;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> n >> m;
    for (int i = 1; i <= m; i++) {
      int t;
      cin >> t >> l[i] >> r[i];
      to[t].push_back(i);
      val.push_back(l[i]);
      val.push_back(r[i]);
    }
    sort(val.begin(), val.end());
    vector<int> tmp;
    for (int i = 0; i <= val.size() - 1; i++)
      if (i == 0 || val[i] != val[i - 1]) tmp.push_back(val[i]);
    val = tmp;
    for (int i = 0; i <= val.size() - 1; i++) {
      mapVal[val[i]] = i;
    }
    for (int i = 1; i <= m; i++) l[i] = mapVal[l[i]], r[i] = mapVal[r[i]];
    build(1, 0, val.size() - 1);
    pair<int, int> ans = mp(n - 1, 1);
    for (int i = 1; i <= n; i++) {
      pair<int, int> cnt = mp(-1, 0);
      for (auto id : to[i]) {
        minimize(cnt, query(1, l[id], r[id]));
      }
      trace[i] = cnt.second;
      cnt.first += i;
      minimize(ans, mp(cnt.first + n - i, i));
      for (auto id : to[i]) {
        update(1, l[id], r[id], mp(cnt.first - i - 1, i));
      }
    }
    cout << ans.first << endl;
    int pos = ans.second;
    for (int i = pos + 1; i <= n; i++) cout << i << " ";
    while (pos != 0) {
      for (int i = trace[pos] + 1; i <= pos - 1; i++) cout << i << " ";
      pos = trace[pos];
    }
  }
  