#include <bits/stdc++.h>
using namespace std;
const int MX = 3e5 + 10;
const int INF = 1e9;
bool vis[MX];
set<int> s;
int cnt[MX], N, M, dp[MX], pre[MX];
vector<pair<int, int> > rg;
vector<int> adj[MX];
void chmax(int& a, int b) {
  if (b > a) a = b;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> M;
  for (int i = 0; i < M; i++) {
    int x, l, r;
    cin >> x >> l >> r;
    rg.push_back({l, x});
    rg.push_back({r + 1, -x});
  }
  sort(rg.begin(), rg.end());
  set<int>::iterator it;
  s.insert(0);
  s.insert(N + 1);
  for (auto p : rg) {
    int id = p.second > 0 ? p.second : -p.second;
    if (p.second > 0) {
      cnt[id]++;
      if (cnt[id] == 1) {
        s.insert(id);
        it = s.find(id);
        it--;
        adj[id].push_back((*it));
        it++;
        it++;
        adj[(*it)].push_back(id);
      }
    } else {
      cnt[id]--;
      if (cnt[id] == 0) s.erase(id);
    }
  }
  for (int i = 1; i <= N + 1; i++) dp[i] = -INF;
  for (int i = 1; i <= N + 1; i++) {
    for (auto j : adj[i]) {
      if (dp[j] + 1 > dp[i]) {
        dp[i] = dp[j] + 1;
        pre[i] = j;
      }
    }
  }
  cout << N + 1 - dp[N + 1] << '\n';
  int curr = N + 1;
  while (curr != 0) {
    vis[curr] = 1;
    curr = pre[curr];
  }
  for (int i = 1; i <= N; i++)
    if (!vis[i]) cout << i << " ";
  cout << '\n';
}

### Prompt

Your task is to create 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;
const int MX = 3e5 + 10;
const int INF = 1e9;
bool vis[MX];
set<int> s;
int cnt[MX], N, M, dp[MX], pre[MX];
vector<pair<int, int> > rg;
vector<int> adj[MX];
void chmax(int& a, int b) {
  if (b > a) a = b;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> N >> M;
  for (int i = 0; i < M; i++) {
    int x, l, r;
    cin >> x >> l >> r;
    rg.push_back({l, x});
    rg.push_back({r + 1, -x});
  }
  sort(rg.begin(), rg.end());
  set<int>::iterator it;
  s.insert(0);
  s.insert(N + 1);
  for (auto p : rg) {
    int id = p.second > 0 ? p.second : -p.second;
    if (p.second > 0) {
      cnt[id]++;
      if (cnt[id] == 1) {
        s.insert(id);
        it = s.find(id);
        it--;
        adj[id].push_back((*it));
        it++;
        it++;
        adj[(*it)].push_back(id);
      }
    } else {
      cnt[id]--;
      if (cnt[id] == 0) s.erase(id);
    }
  }
  for (int i = 1; i <= N + 1; i++) dp[i] = -INF;
  for (int i = 1; i <= N + 1; i++) {
    for (auto j : adj[i]) {
      if (dp[j] + 1 > dp[i]) {
        dp[i] = dp[j] + 1;
        pre[i] = j;
      }
    }
  }
  cout << N + 1 - dp[N + 1] << '\n';
  int curr = N + 1;
  while (curr != 0) {
    vis[curr] = 1;
    curr = pre[curr];
  }
  for (int i = 1; i <= N; i++)
    if (!vis[i]) cout << i << " ";
  cout << '\n';
}

```