#include <bits/stdc++.h>
using namespace std;
const long long INF = (long long)1 << 62;
const long long MOD = 1e9 + 7;
const int iINF = 1 << 30;
const double PI = 3.14159265359;
struct intv {
  int i, tmx, tpe;
};
bool cmp(intv a, intv b) {
  if (a.tmx == b.tmx) {
    if (a.tpe == b.tpe) return a.i < b.i;
    return a.tpe < b.tpe;
  }
  return a.tmx < b.tmx;
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<intv> a;
  for (int i = 0; i < m; i++) {
    int x, l, r;
    cin >> x >> l >> r;
    a.push_back(intv{x - 1, l, 0});
    a.push_back(intv{x - 1, r, 1});
  }
  sort(a.begin(), a.end(), cmp);
  vector<vector<int> > adj(n);
  map<int, int> active;
  for (int i = 0; i < (int)a.size(); i++) {
    if (a[i].tpe == 0) {
      if (active.find(a[i].i) == active.end()) active[a[i].i] = 0;
      active[a[i].i]++;
      auto it = active.lower_bound(a[i].i);
      it++;
      if (it != active.end()) adj[a[i].i].push_back(it->first);
      it--;
      if (it != active.begin()) {
        it--;
        adj[it->first].push_back(a[i].i);
      }
    } else {
      if (active[a[i].i] == 1) {
        auto it = active.lower_bound(a[i].i);
        it++;
        int u = -1, v = -1;
        if (it != active.end()) v = it->first;
        it--;
        if (it != active.begin()) {
          it--;
          u = it->first;
        }
        if (u != -1 && v != -1) adj[u].push_back(v);
      }
      active[a[i].i]--;
      if (active[a[i].i] == 0) active.erase(active.find(a[i].i));
    }
  }
  vector<int> dp(n, 1);
  int strt = 0;
  for (int u = n - 1; u >= 0; u--) {
    for (int v : adj[u]) dp[u] = max(dp[u], dp[v] + 1);
    if (dp[u] > dp[strt]) strt = u;
  }
  set<int> taken;
  int curr = strt;
  taken.insert(curr);
  while ((int)taken.size() != dp[strt]) {
    int maxi = adj[curr][0];
    for (int v : adj[curr])
      if (dp[v] > dp[maxi]) maxi = v;
    curr = maxi;
    taken.insert(curr);
  }
  cout << n - taken.size() << "\n";
  for (int i = 0; i < n; i++) {
    if (taken.find(i) == taken.end()) cout << i + 1 << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 long long INF = (long long)1 << 62;
const long long MOD = 1e9 + 7;
const int iINF = 1 << 30;
const double PI = 3.14159265359;
struct intv {
  int i, tmx, tpe;
};
bool cmp(intv a, intv b) {
  if (a.tmx == b.tmx) {
    if (a.tpe == b.tpe) return a.i < b.i;
    return a.tpe < b.tpe;
  }
  return a.tmx < b.tmx;
}
void solve() {
  int n, m;
  cin >> n >> m;
  vector<intv> a;
  for (int i = 0; i < m; i++) {
    int x, l, r;
    cin >> x >> l >> r;
    a.push_back(intv{x - 1, l, 0});
    a.push_back(intv{x - 1, r, 1});
  }
  sort(a.begin(), a.end(), cmp);
  vector<vector<int> > adj(n);
  map<int, int> active;
  for (int i = 0; i < (int)a.size(); i++) {
    if (a[i].tpe == 0) {
      if (active.find(a[i].i) == active.end()) active[a[i].i] = 0;
      active[a[i].i]++;
      auto it = active.lower_bound(a[i].i);
      it++;
      if (it != active.end()) adj[a[i].i].push_back(it->first);
      it--;
      if (it != active.begin()) {
        it--;
        adj[it->first].push_back(a[i].i);
      }
    } else {
      if (active[a[i].i] == 1) {
        auto it = active.lower_bound(a[i].i);
        it++;
        int u = -1, v = -1;
        if (it != active.end()) v = it->first;
        it--;
        if (it != active.begin()) {
          it--;
          u = it->first;
        }
        if (u != -1 && v != -1) adj[u].push_back(v);
      }
      active[a[i].i]--;
      if (active[a[i].i] == 0) active.erase(active.find(a[i].i));
    }
  }
  vector<int> dp(n, 1);
  int strt = 0;
  for (int u = n - 1; u >= 0; u--) {
    for (int v : adj[u]) dp[u] = max(dp[u], dp[v] + 1);
    if (dp[u] > dp[strt]) strt = u;
  }
  set<int> taken;
  int curr = strt;
  taken.insert(curr);
  while ((int)taken.size() != dp[strt]) {
    int maxi = adj[curr][0];
    for (int v : adj[curr])
      if (dp[v] > dp[maxi]) maxi = v;
    curr = maxi;
    taken.insert(curr);
  }
  cout << n - taken.size() << "\n";
  for (int i = 0; i < n; i++) {
    if (taken.find(i) == taken.end()) cout << i + 1 << " ";
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```