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;
  template <int maxn>
  struct lazy {
    pair<int, int> tree[maxn * 4 + 5], pend[maxn * 4 + 5];
    lazy() {
      for (int i = 0; i < maxn * 4 + 3; ++i)
        tree[i] = {1000000000, 1000000000}, pend[i] = {1000000000, 1000000000};
    }
    void propagate(int cur, int ss, int se) {
      if (pend[cur].first != 1000000000) {
        tree[cur] = min(tree[cur], pend[cur]);
        if (ss != se) {
          pend[cur * 2 + 1] = min(pend[cur * 2 + 1], pend[cur]);
          pend[cur * 2 + 2] = min(pend[cur * 2 + 2], pend[cur]);
        }
        pend[cur].first = 1000000000;
      }
    }
    void addUtil(int cur, int ss, int se, int l, int r, pair<int, int> val) {
      propagate(cur, ss, se);
      if (ss > se || ss > r || se < l) return;
      if (ss >= l && se <= r) {
        pend[cur] = val;
        propagate(cur, ss, se);
        return;
      }
      addUtil(cur * 2 + 1, ss, (ss + se) / 2, l, r, val);
      addUtil(cur * 2 + 2, (ss + se) / 2 + 1, se, l, r, val);
      tree[cur] = min(tree[cur * 2 + 1], tree[cur * 2 + 2]);
    }
    void add(int l, int r, pair<int, int> val) {
      addUtil(0, 0, maxn - 1, l, r, val);
    }
    pair<int, int> getUtil(int ss, int se, int l, int r, int cur) {
      propagate(cur, ss, se);
      if (ss > se || ss > r || se < l) return {1000000000, 1000000000};
      if (ss >= l && se <= r) {
        return tree[cur];
      }
      return min(getUtil(ss, (ss + se) / 2, l, r, cur * 2 + 1),
                 getUtil((ss + se) / 2 + 1, se, l, r, cur * 2 + 2));
    }
    pair<int, int> get(int l, int r) { return getUtil(0, maxn - 1, l, r, 0); }
  };
  lazy<600005> dps;
  int dp[300005];
  vector<pair<int, int>> seg[300005];
  int back[300005];
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < m; ++i) {
      int j, l, r;
      cin >> j >> l >> r;
      --j;
      seg[j].push_back({l, r});
    }
    for (int i = 0; i < n; ++i) sort(seg[i].begin(), seg[i].end());
    set<int> all;
    for (int i = 0; i < n; ++i) {
      for (auto j : seg[i]) all.insert(j.first), all.insert(j.second);
    }
    map<int, int> indx;
    int k = 0;
    for (auto i : all) {
      indx[i] = k++;
    }
    for (int i = 0; i < n; ++i) {
      for (auto& j : seg[i]) {
        j.first = indx[j.first];
        j.second = indx[j.second];
      }
    }
    dp[n - 1] = 0;
    back[n - 1] = -1;
    for (auto j : seg[n - 1]) {
      auto [l, r] = j;
      dps.add(l, r, {n - 1, n - 1});
    }
    for (int i = n - 2; i >= 0; --i) {
      int& res = dp[i];
      res = n - i - 1;
      back[i] = -1;
      for (auto j : seg[i]) {
        auto [l, r] = j;
        if (res > dps.get(l, r).first - (i + 1)) {
          back[i] = dps.get(l, r).second;
        }
        res = min(res, dps.get(l, r).first - (i + 1));
      }
      for (auto j : seg[i]) {
        auto [l, r] = j;
        dps.add(l, r, {res + i, i});
      }
    }
    int ans = 1000000000;
    for (int i = 0; i < n; ++i) ans = min(ans, dp[i] + i);
    assert(ans != 1000000000);
    cout << ans << endl;
    for (int i = 0; i < n; ++i) {
      if (ans == dp[i] + i) {
        int cur = i;
        vector<int> all;
        while (back[cur] != -1) {
          all.push_back(cur);
          cur = back[cur];
        }
        all.push_back(cur);
        for (int j = 0; j < i; ++j) cout << j + 1 << ' ';
        for (int j = 0; j + 1 < all.size(); ++j) {
          for (int k = all[j] + 1; k < all[j + 1]; ++k) {
            cout << k + 1 << ' ';
          }
        }
        if (!all.empty())
          for (int j = all.back() + 1; j < n; ++j) cout << j + 1 << ' ';
        cout << endl;
        return 0;
      }
    }
    assert(0);
  }
  