#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  vector<int> cover(n);
  vector<tuple<int, int, int>> endpoints;
  set<int> s{-1, n};
  for (int i = 0; i < m; i++) {
    int x, l, r;
    cin >> x >> l >> r;
    x--;
    l--;
    endpoints.emplace_back(l, x, -1);
    endpoints.emplace_back(r, x, 1);
  }
  sort(endpoints.begin(), endpoints.end());
  vector<pair<int, int>> edges;
  vector<int> add, del;
  for (int l = 0, r; l < 2 * m; l = r) {
    r = l;
    while (r < 2 * m && get<0>(endpoints[l]) == get<0>(endpoints[r])) r++;
    add.clear();
    del.clear();
    for (int i = l; i < r; i++) {
      auto [v, x, t] = endpoints[i];
      if (t == -1) {
        if (cover[x] == 0) add.push_back(x);
        cover[x]++;
      } else {
        cover[x]--;
        if (cover[x] == 0) del.push_back(x);
      }
    }
    for (int x : del) s.erase(x);
    for (int x : add) {
      auto it = s.insert(x).first;
      edges.emplace_back(*prev(it), x);
      edges.emplace_back(x, *next(it));
    }
  }
  sort(edges.begin(), edges.end());
  vector<int> dp(n + 2), g(n + 2);
  for (auto [u, v] : edges) {
    if (dp[u + 1] + 1 > dp[v + 1]) {
      dp[v + 1] = dp[u + 1] + 1;
      g[v + 1] = u + 1;
    }
  }
  int res = n + 1 - dp[n + 1];
  cout << res << "\n";
  vector<int> ans;
  for (int i = n + 1; i > 0; i = g[i]) {
    for (int j = g[i] + 1; j < i; j++) {
      ans.push_back(j);
    }
  }
  for (int i = 0; i < res; i++) cout << ans[i] << " \n"[i == res - 1];
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  vector<int> cover(n);
  vector<tuple<int, int, int>> endpoints;
  set<int> s{-1, n};
  for (int i = 0; i < m; i++) {
    int x, l, r;
    cin >> x >> l >> r;
    x--;
    l--;
    endpoints.emplace_back(l, x, -1);
    endpoints.emplace_back(r, x, 1);
  }
  sort(endpoints.begin(), endpoints.end());
  vector<pair<int, int>> edges;
  vector<int> add, del;
  for (int l = 0, r; l < 2 * m; l = r) {
    r = l;
    while (r < 2 * m && get<0>(endpoints[l]) == get<0>(endpoints[r])) r++;
    add.clear();
    del.clear();
    for (int i = l; i < r; i++) {
      auto [v, x, t] = endpoints[i];
      if (t == -1) {
        if (cover[x] == 0) add.push_back(x);
        cover[x]++;
      } else {
        cover[x]--;
        if (cover[x] == 0) del.push_back(x);
      }
    }
    for (int x : del) s.erase(x);
    for (int x : add) {
      auto it = s.insert(x).first;
      edges.emplace_back(*prev(it), x);
      edges.emplace_back(x, *next(it));
    }
  }
  sort(edges.begin(), edges.end());
  vector<int> dp(n + 2), g(n + 2);
  for (auto [u, v] : edges) {
    if (dp[u + 1] + 1 > dp[v + 1]) {
      dp[v + 1] = dp[u + 1] + 1;
      g[v + 1] = u + 1;
    }
  }
  int res = n + 1 - dp[n + 1];
  cout << res << "\n";
  vector<int> ans;
  for (int i = n + 1; i > 0; i = g[i]) {
    for (int j = g[i] + 1; j < i; j++) {
      ans.push_back(j);
    }
  }
  for (int i = 0; i < res; i++) cout << ans[i] << " \n"[i == res - 1];
  return 0;
}

```