#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<pair<int, int>> endpoint;
  for (int i = 0; i < m; i++) {
    int id, l, r;
    cin >> id >> l >> r;
    endpoint.emplace_back(l, id);
    endpoint.emplace_back(r + 1, (~id));
  }
  sort(endpoint.begin(), endpoint.end());
  vector<vector<int>> G(n + 1);
  map<int, int> mp;
  int beg = 0;
  while (beg < (int)endpoint.size()) {
    int end = beg;
    while (end + 1 < (int)endpoint.size() &&
           endpoint[end + 1].first == endpoint[end].first) {
      ++end;
    }
    for (int i = beg; i <= end; i++) {
      int id = endpoint[i].second;
      if (id >= 0) {
        auto it = mp.lower_bound(id);
        if (it != mp.end() && it->first == id) {
          it->second += 1;
          continue;
        }
        if (it != mp.begin()) {
          int x = prev(it)->first;
          G[x].push_back(id);
        }
        if (it != mp.end()) {
          int y = it->first;
          G[id].push_back(y);
        }
        mp[id] = 1;
      } else {
        id = ~id;
        auto it = mp.find(id);
        assert(it != mp.end());
        if (it->second > 1) {
          it->second -= 1;
          continue;
        }
        if (it != mp.begin() && next(it) != mp.end()) {
          int x = prev(it)->first;
          int y = next(it)->first;
          G[x].push_back(y);
        }
        mp.erase(it);
      }
    }
    beg = end + 1;
  }
  vector<int> dp(n + 1, 1);
  vector<int> pa(n + 1, -1);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < (int)G[i].size(); j++) {
      if (dp[i] + 1 > dp[G[i][j]]) {
        dp[G[i][j]] = dp[i] + 1;
        pa[G[i][j]] = i;
      }
    }
  }
  int pos = (int)(max_element(dp.begin() + 1, dp.end()) - dp.begin());
  int ans = n;
  vector<bool> remain(n + 1, 0);
  while (pos > 0) {
    remain[pos] = 1;
    ans--;
    pos = pa[pos];
  }
  cout << ans << '\n';
  for (int i = 1; i <= n; i++) {
    if (!remain[i]) {
      cout << i << " ";
    }
  }
  cout << '\n';
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<pair<int, int>> endpoint;
  for (int i = 0; i < m; i++) {
    int id, l, r;
    cin >> id >> l >> r;
    endpoint.emplace_back(l, id);
    endpoint.emplace_back(r + 1, (~id));
  }
  sort(endpoint.begin(), endpoint.end());
  vector<vector<int>> G(n + 1);
  map<int, int> mp;
  int beg = 0;
  while (beg < (int)endpoint.size()) {
    int end = beg;
    while (end + 1 < (int)endpoint.size() &&
           endpoint[end + 1].first == endpoint[end].first) {
      ++end;
    }
    for (int i = beg; i <= end; i++) {
      int id = endpoint[i].second;
      if (id >= 0) {
        auto it = mp.lower_bound(id);
        if (it != mp.end() && it->first == id) {
          it->second += 1;
          continue;
        }
        if (it != mp.begin()) {
          int x = prev(it)->first;
          G[x].push_back(id);
        }
        if (it != mp.end()) {
          int y = it->first;
          G[id].push_back(y);
        }
        mp[id] = 1;
      } else {
        id = ~id;
        auto it = mp.find(id);
        assert(it != mp.end());
        if (it->second > 1) {
          it->second -= 1;
          continue;
        }
        if (it != mp.begin() && next(it) != mp.end()) {
          int x = prev(it)->first;
          int y = next(it)->first;
          G[x].push_back(y);
        }
        mp.erase(it);
      }
    }
    beg = end + 1;
  }
  vector<int> dp(n + 1, 1);
  vector<int> pa(n + 1, -1);
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < (int)G[i].size(); j++) {
      if (dp[i] + 1 > dp[G[i][j]]) {
        dp[G[i][j]] = dp[i] + 1;
        pa[G[i][j]] = i;
      }
    }
  }
  int pos = (int)(max_element(dp.begin() + 1, dp.end()) - dp.begin());
  int ans = n;
  vector<bool> remain(n + 1, 0);
  while (pos > 0) {
    remain[pos] = 1;
    ans--;
    pos = pa[pos];
  }
  cout << ans << '\n';
  for (int i = 1; i <= n; i++) {
    if (!remain[i]) {
      cout << i << " ";
    }
  }
  cout << '\n';
  return 0;
}

```