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 data {
    long long id, l, r;
    bool operator<(const data &a) const { return id < a.id; }
  };
  const long long N = 6e5 + 10;
  long long n;
  long long m;
  data a[N];
  struct Interval_Tree {
    pair<long long, long long> node[4 * N];
    pair<long long, long long> lazy[4 * N];
    Interval_Tree() {}
    void Down(long long i) {
      if (node[2 * i].first < lazy[i].first) {
        node[2 * i] = lazy[i];
      }
      if (node[2 * i + 1].first < lazy[i].first) {
        node[2 * i + 1] = lazy[i];
      }
      if (lazy[2 * i].first < lazy[i].first) {
        lazy[2 * i] = lazy[i];
      }
      if (lazy[2 * i + 1].first < lazy[i].first) {
        lazy[2 * i + 1] = lazy[i];
      }
      lazy[i] = {0, 0};
    }
    void update(long long i, long long l, long long r, long long x, long long y,
                long long val, long long id) {
      if (l > y || x > r) {
        return;
      }
      if (x <= l && r <= y) {
        if (node[i].first < val) {
          node[i] = {val, id};
        }
        if (lazy[i].first < val) {
          lazy[i] = {val, id};
        }
        return;
      }
      Down(i);
      long long mid = (l + r) / 2;
      update(2 * i, l, mid, x, y, val, id);
      update(2 * i + 1, mid + 1, r, x, y, val, id);
      if (node[2 * i].first < node[2 * i + 1].first) {
        node[i] = node[2 * i + 1];
      } else {
        node[i] = node[2 * i];
      }
    }
    pair<long long, long long> get(long long i, long long l, long long r,
                                   long long x, long long y) {
      if (l > y || x > r) {
        return {0, 0};
      }
      if (x <= l && r <= y) {
        return node[i];
      }
      Down(i);
      long long mid = (l + r) / 2;
      pair<long long, long long> tmp1 = get(2 * i, l, mid, x, y);
      pair<long long, long long> tmp2 = get(2 * i + 1, mid + 1, r, x, y);
      if (tmp1.first < tmp2.first) {
        return tmp2;
      } else {
        return tmp1;
      }
    }
  } IT;
  long long maxVal;
  void compress() {
    vector<long long> rar;
    for (long long i = 1; i <= m; ++i) {
      rar.push_back(a[i].l);
      rar.push_back(a[i].r);
    }
    sort(rar.begin(), rar.end());
    rar.resize(unique(rar.begin(), rar.end()) - rar.begin());
    for (long long i = 1; i <= m; ++i) {
      a[i].l = lower_bound(rar.begin(), rar.end(), a[i].l) - rar.begin() + 1;
      a[i].r = lower_bound(rar.begin(), rar.end(), a[i].r) - rar.begin() + 1;
    }
    maxVal = (long long)rar.size() + 1;
  }
  vector<long long> V;
  bool mark[N];
  pair<long long, long long> dp[N];
  void sol() {
    compress();
    sort(a + 1, a + m + 1);
    IT.update(1, 1, maxVal, 1, maxVal, 0, 0);
    for (long long l = 1; l <= m; ++l) {
      long long r = l;
      while (r <= m && a[r].id == a[l].id) {
        ++r;
      }
      --r;
      for (long long i = l; i <= r; ++i) {
        long long id = a[i].id;
        pair<long long, long long> tmp = IT.get(1, 1, maxVal, a[i].l, a[i].r);
        if (dp[id].first < tmp.first + 1) {
          dp[id] = {tmp.first + 1, tmp.second};
        }
      }
      for (long long i = l; i <= r; ++i) {
        long long id = a[i].id;
        IT.update(1, 1, maxVal, a[i].l, a[i].r, dp[id].first, id);
      }
      l = r;
    }
    long long res = 0;
    long long save = -1;
    for (long long i = 1; i <= n; ++i) {
      if (res < dp[i].first) {
        res = dp[i].first;
        save = i;
      }
    }
    while (save > 0) {
      mark[save] = true;
      save = dp[save].second;
    }
    for (long long i = 1; i <= n; ++i) {
      if (!mark[i]) {
        V.push_back(i);
      }
    }
    cout << V.size() << '\n';
    for (long long r : V) {
      cout << r << " ";
    }
    cout << '\n';
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    if (fopen("input.txt", "r")) {
      freopen("input.txt", "r", stdin);
      freopen("output.txt", "w", stdout);
    }
    cin >> n >> m;
    for (long long i = 1; i <= m; ++i) {
      long long id, l, r;
      cin >> id >> l >> r;
      a[i] = {id, l, r};
    }
    sol();
    return 0;
  }
  