#include <bits/stdc++.h>
using namespace std;
struct Node {
  Node* left = nullptr;
  Node* right = nullptr;
  pair<int, int> mmax = {0, 0};
  pair<int, int> assign = {-1, -1};
  Node() {}
  Node(pair<int, int> mm, pair<int, int> a)
      : left(nullptr), right(nullptr), mmax(mm), assign(a) {}
  void push() {
    if (assign.second != -1) {
      left->assign = assign;
      right->assign = assign;
    }
    assign = {-1, -1};
  }
  void create_children() {
    if (!left) {
      left = new Node();
    }
    if (!right) {
      right = new Node();
    }
  }
  void update() {
    mmax = max((left->assign.second != -1 ? left->assign : left->mmax),
               (right->assign.second != -1 ? right->assign : right->mmax));
  }
  pair<int, int> qmax(int lq, int rq, int lb = 0, int rb = 7e5 + 1) {
    if (lq <= lb && rb <= rq) {
      return (assign.second != -1 ? assign : mmax);
    } else if (lq >= rb || rq <= lb) {
      return {0, 0};
    }
    create_children();
    push();
    int mid = (lb + rb) >> 1;
    auto res = max(left->qmax(lq, rq, lb, mid), right->qmax(lq, rq, mid, rb));
    update();
    return res;
  }
  void qassign(int lq, int rq, pair<int, int> val, int lb = 0,
               int rb = 7e5 + 1) {
    if (lq <= lb && rb <= rq) {
      assign = val;
      return;
    } else if (lq >= rb || rq <= lb) {
      return;
    }
    create_children();
    push();
    int mid = (lb + rb) >> 1;
    left->qassign(lq, rq, val, lb, mid);
    right->qassign(lq, rq, val, mid, rb);
    update();
  }
  ~Node() {
    delete left;
    delete right;
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<vector<pair<int, int>>> ranges(1 + n);
  vector<int> coord;
  for (int i = 0; i < m; ++i) {
    int ind, left, right;
    cin >> ind >> left >> right;
    ++right;
    ranges[ind].push_back({left, right});
    coord.push_back(left);
    coord.push_back(right);
  }
  sort(coord.begin(), coord.end());
  coord.resize(unique(coord.begin(), coord.end()) - coord.begin());
  for (int i = 1; i <= n; ++i) {
    for (pair<int, int>& r : ranges[i]) {
      r.first =
          lower_bound(coord.begin(), coord.end(), r.first) - coord.begin();
      r.second =
          lower_bound(coord.begin(), coord.end(), r.second) - coord.begin();
    }
  }
  vector<int> dp(1 + n, 0);
  vector<int> prev(1 + n, -1);
  Node* root = new Node();
  for (int i = 1; i <= n; ++i) {
    pair<int, int> mm = {0, 0};
    for (pair<int, int> rr : ranges[i]) {
      mm = max(mm, root->qmax(rr.first, rr.second));
    }
    dp[i] = mm.first + 1;
    prev[i] = mm.second;
    for (pair<int, int> rr : ranges[i]) {
      root->qassign(rr.first, rr.second, {mm.first + 1, i});
    }
  }
  int ind = max_element(dp.begin(), dp.end()) - dp.begin();
  cout << n - dp[ind] << "\n";
  vector<bool> need(1 + n, false);
  while (ind != 0) {
    need[ind] = true;
    ind = prev[ind];
  }
  for (int i = 1; i <= n; ++i) {
    if (!need[i]) {
      cout << i << " ";
    }
  }
}

### Prompt

Generate a CPP solution to the following problem:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
struct Node {
  Node* left = nullptr;
  Node* right = nullptr;
  pair<int, int> mmax = {0, 0};
  pair<int, int> assign = {-1, -1};
  Node() {}
  Node(pair<int, int> mm, pair<int, int> a)
      : left(nullptr), right(nullptr), mmax(mm), assign(a) {}
  void push() {
    if (assign.second != -1) {
      left->assign = assign;
      right->assign = assign;
    }
    assign = {-1, -1};
  }
  void create_children() {
    if (!left) {
      left = new Node();
    }
    if (!right) {
      right = new Node();
    }
  }
  void update() {
    mmax = max((left->assign.second != -1 ? left->assign : left->mmax),
               (right->assign.second != -1 ? right->assign : right->mmax));
  }
  pair<int, int> qmax(int lq, int rq, int lb = 0, int rb = 7e5 + 1) {
    if (lq <= lb && rb <= rq) {
      return (assign.second != -1 ? assign : mmax);
    } else if (lq >= rb || rq <= lb) {
      return {0, 0};
    }
    create_children();
    push();
    int mid = (lb + rb) >> 1;
    auto res = max(left->qmax(lq, rq, lb, mid), right->qmax(lq, rq, mid, rb));
    update();
    return res;
  }
  void qassign(int lq, int rq, pair<int, int> val, int lb = 0,
               int rb = 7e5 + 1) {
    if (lq <= lb && rb <= rq) {
      assign = val;
      return;
    } else if (lq >= rb || rq <= lb) {
      return;
    }
    create_children();
    push();
    int mid = (lb + rb) >> 1;
    left->qassign(lq, rq, val, lb, mid);
    right->qassign(lq, rq, val, mid, rb);
    update();
  }
  ~Node() {
    delete left;
    delete right;
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<vector<pair<int, int>>> ranges(1 + n);
  vector<int> coord;
  for (int i = 0; i < m; ++i) {
    int ind, left, right;
    cin >> ind >> left >> right;
    ++right;
    ranges[ind].push_back({left, right});
    coord.push_back(left);
    coord.push_back(right);
  }
  sort(coord.begin(), coord.end());
  coord.resize(unique(coord.begin(), coord.end()) - coord.begin());
  for (int i = 1; i <= n; ++i) {
    for (pair<int, int>& r : ranges[i]) {
      r.first =
          lower_bound(coord.begin(), coord.end(), r.first) - coord.begin();
      r.second =
          lower_bound(coord.begin(), coord.end(), r.second) - coord.begin();
    }
  }
  vector<int> dp(1 + n, 0);
  vector<int> prev(1 + n, -1);
  Node* root = new Node();
  for (int i = 1; i <= n; ++i) {
    pair<int, int> mm = {0, 0};
    for (pair<int, int> rr : ranges[i]) {
      mm = max(mm, root->qmax(rr.first, rr.second));
    }
    dp[i] = mm.first + 1;
    prev[i] = mm.second;
    for (pair<int, int> rr : ranges[i]) {
      root->qassign(rr.first, rr.second, {mm.first + 1, i});
    }
  }
  int ind = max_element(dp.begin(), dp.end()) - dp.begin();
  cout << n - dp[ind] << "\n";
  vector<bool> need(1 + n, false);
  while (ind != 0) {
    need[ind] = true;
    ind = prev[ind];
  }
  for (int i = 1; i <= n; ++i) {
    if (!need[i]) {
      cout << i << " ";
    }
  }
}

```