#include <bits/stdc++.h>
using namespace std;
struct point {
  int id;
  bool ok;
  int row;
  bool operator<(const point &p) const {
    return (id == p.id) ? ok < p.ok : id < p.id;
  }
};
point a[600100];
vector<int> edge[300100];
int dp[300100];
int truoc[300100];
bool ok[300100];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    a[i << 1] = {y, 1, x};
    a[i << 1 | 1] = {z + 1, 0, x};
  }
  sort(a, a + 2 * m);
  multiset<int, greater<int> > b;
  multiset<int> c;
  for (int i = 0; i < 2 * m; i++) {
    if (a[i].ok) {
      if (b.upper_bound(a[i].row) != b.end()) {
        edge[*b.upper_bound(a[i].row)].push_back(a[i].row);
      }
      if (c.upper_bound(a[i].row) != c.end()) {
        edge[a[i].row].push_back(*c.upper_bound(a[i].row));
      }
      b.insert(a[i].row);
      c.insert(a[i].row);
    } else {
      b.erase(b.find(a[i].row));
      c.erase(c.find(a[i].row));
    }
  }
  for (int i = 1; i < n; i++) {
    dp[i] = max(dp[i], 1);
    for (auto x : edge[i]) {
      if (dp[x] < dp[i] + 1) {
        dp[x] = dp[i] + 1;
        truoc[x] = i;
      }
    }
  }
  int max1 = -1;
  int id;
  for (int i = 1; i <= n; i++) {
    if (dp[i] > max1) {
      max1 = dp[i];
      id = i;
    }
  }
  cout << n - max1 << '\n';
  int cur = id;
  while (cur != 0) {
    ok[cur] = 1;
    cur = truoc[cur];
  }
  for (int i = 1; i <= n; i++) {
    if (!ok[i]) {
      cout << i << ' ';
    }
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
struct point {
  int id;
  bool ok;
  int row;
  bool operator<(const point &p) const {
    return (id == p.id) ? ok < p.ok : id < p.id;
  }
};
point a[600100];
vector<int> edge[300100];
int dp[300100];
int truoc[300100];
bool ok[300100];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int x, y, z;
    cin >> x >> y >> z;
    a[i << 1] = {y, 1, x};
    a[i << 1 | 1] = {z + 1, 0, x};
  }
  sort(a, a + 2 * m);
  multiset<int, greater<int> > b;
  multiset<int> c;
  for (int i = 0; i < 2 * m; i++) {
    if (a[i].ok) {
      if (b.upper_bound(a[i].row) != b.end()) {
        edge[*b.upper_bound(a[i].row)].push_back(a[i].row);
      }
      if (c.upper_bound(a[i].row) != c.end()) {
        edge[a[i].row].push_back(*c.upper_bound(a[i].row));
      }
      b.insert(a[i].row);
      c.insert(a[i].row);
    } else {
      b.erase(b.find(a[i].row));
      c.erase(c.find(a[i].row));
    }
  }
  for (int i = 1; i < n; i++) {
    dp[i] = max(dp[i], 1);
    for (auto x : edge[i]) {
      if (dp[x] < dp[i] + 1) {
        dp[x] = dp[i] + 1;
        truoc[x] = i;
      }
    }
  }
  int max1 = -1;
  int id;
  for (int i = 1; i <= n; i++) {
    if (dp[i] > max1) {
      max1 = dp[i];
      id = i;
    }
  }
  cout << n - max1 << '\n';
  int cur = id;
  while (cur != 0) {
    ok[cur] = 1;
    cur = truoc[cur];
  }
  for (int i = 1; i <= n; i++) {
    if (!ok[i]) {
      cout << i << ' ';
    }
  }
  return 0;
}

```