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 pair<long long, long long> NO_PROP = {-1, -1};
  class SegTree {
   public:
    pair<long long, long long> maxi;
    pair<long long, long long> toProp = NO_PROP;
    SegTree *left, *right;
    long long lx, rx;
    SegTree(long long l, long long r) {
      lx = l;
      rx = r;
      maxi = {0, -1};
      toProp = NO_PROP;
      left = NULL, right = NULL;
    }
    SegTree* getLeft() {
      assert(lx != rx);
      long long mid = (lx + rx) / 2;
      if (!left) left = new SegTree(lx, mid);
      return left;
    }
    SegTree* getRight() {
      assert(lx != rx);
      long long mid = (lx + rx) / 2;
      if (!right) right = new SegTree(mid + 1, rx);
      return right;
    }
    void prop() {
      if (lx != rx && toProp != NO_PROP) {
        getLeft()->toProp = toProp;
        getRight()->toProp = toProp;
        if (getLeft()->maxi.first < toProp.first) {
          getLeft()->maxi = toProp;
        }
        if (getRight()->maxi.first < toProp.first) {
          getRight()->maxi = toProp;
        }
        toProp = NO_PROP;
      }
    }
    void rangeUpdate(long long l, long long r, pair<long long, long long> val) {
      this->prop();
      if (rx < l || r < lx) return;
      if (l <= lx && rx <= r) {
        if (maxi.first <= val.first) {
          maxi = val;
        }
        toProp = val;
        return;
      }
      getLeft()->rangeUpdate(l, r, val);
      getRight()->rangeUpdate(l, r, val);
      maxi = max(getLeft()->maxi, getRight()->maxi);
    }
    pair<long long, long long> rangeVal(long long l, long long r) {
      this->prop();
      if (rx < l || r < lx) return {0, -1};
      if (l <= lx && rx <= r) {
        return maxi;
      }
      auto a = getLeft()->rangeVal(l, r);
      auto b = getRight()->rangeVal(l, r);
      return max(a, b);
    }
  };
  void pilot() {
    long long n, m;
    cin >> n >> m;
    vector<vector<pair<long long, long long>>> a(n);
    vector<long long> tmp;
    for (long long i = 0; i < m; i++) {
      long long row, l, r;
      cin >> row >> l >> r;
      tmp.push_back(l);
      tmp.push_back(r);
      a[row - 1].push_back({l, r});
    }
    sort(tmp.begin(), tmp.end());
    map<long long, long long> actualToCompressedCoords;
    long long siz = tmp.size();
    for (int i = 0; i < siz;) {
      int j = i;
      while (j < siz && tmp[i] == tmp[j]) j++;
      actualToCompressedCoords[tmp[i]] = i;
      i = j;
    }
    auto segTree = new SegTree(0, 1e9);
    vector<long long> dp(n), parent(n, -1);
    vector<long long> ans;
    for (long long i = 0; i < n; i++) {
      pair<long long, long long> val = {0, -1};
      for (auto& seg : a[i]) {
        long long l = actualToCompressedCoords[seg.first];
        long long r = actualToCompressedCoords[seg.second];
        auto tmp = segTree->rangeVal(l, r);
        val = max(val, tmp);
      }
      dp[i] = val.first + 1;
      parent[i] = val.second;
      for (auto& seg : a[i]) {
        long long l = actualToCompressedCoords[seg.first];
        long long r = actualToCompressedCoords[seg.second];
        segTree->rangeUpdate(l, r, {val.first + 1, i});
      }
    }
    int maxiIndx = 0;
    for (int i = 0; i < n; i++) {
      if (dp[i] > dp[maxiIndx]) {
        maxiIndx = i;
      }
    }
    vector<long long> visited(n);
    while (parent[maxiIndx] != -1) {
      visited[maxiIndx] = 1;
      maxiIndx = parent[maxiIndx];
    }
    visited[maxiIndx] = 1;
    for (int i = 0; i < n; i++) {
      if (!visited[i]) ans.push_back(i + 1);
    }
    cout << ans.size() << "\n";
    for (int x : ans) cout << x << " ";
    cout << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long tt = 1;
    for (long long i = 0; i < tt; i++) {
      pilot();
    }
    return 0;
  }
  