#include <bits/stdc++.h>
using namespace std;
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << ' ' << H;
  debug_out(T...);
}
const int N = 1e6 + 5;
vector<pair<int, int>> v[N];
int vis[N], las[N], dp[N], b[N], tot;
pair<int, int> tag[N * 4], val[N * 4];
void update(int l, int r, int rt, int a, int b, pair<int, int> p) {
  if (l >= a && r <= b) {
    if (tag[rt] < p) tag[rt] = p;
    val[rt] = max(val[rt], p);
    return;
  }
  int m = l + r >> 1;
  if (a <= m) update(l, m, rt << 1, a, b, p);
  if (b > m) update(m + 1, r, rt << 1 | 1, a, b, p);
  val[rt] = max(val[rt << 1], val[rt << 1 | 1]);
}
pair<int, int> query(int l, int r, int rt, int a, int b) {
  if (l >= a && r <= b) return val[rt];
  int m = l + r >> 1;
  pair<int, int> ans = tag[rt];
  if (a <= m) ans = max(ans, query(l, m, rt << 1, a, b));
  if (b > m) ans = max(ans, query(m + 1, r, rt << 1 | 1, a, b));
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1, j, l, r; i <= m; i++) {
    cin >> j >> l >> r;
    v[j].emplace_back(l, r);
    b[++tot] = l, b[++tot] = r;
  }
  sort(b + 1, b + tot + 1);
  tot = unique(b + 1, b + tot + 1) - b - 1;
  int mx = 0, p = 0;
  for (int i = 1; i <= n; i++) {
    pair<int, int> tmp = {0, 0};
    for (auto& e : v[i]) {
      int l = lower_bound(b + 1, b + tot + 1, e.first) - b;
      int r = lower_bound(b + 1, b + tot + 1, e.second) - b;
      auto pi = query(1, tot, 1, l, r);
      tmp = max(tmp, pi);
    }
    dp[i] = tmp.first + 1, las[i] = tmp.second;
    for (auto& e : v[i]) {
      int l = lower_bound(b + 1, b + tot + 1, e.first) - b;
      int r = lower_bound(b + 1, b + tot + 1, e.second) - b;
      update(1, tot, 1, l, r, {dp[i], i});
    }
    if (mx < dp[i]) {
      mx = dp[i];
      p = i;
    }
  }
  while (p) {
    vis[p] = 1;
    p = las[p];
  }
  vector<int> ans;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) ans.push_back(i);
  cout << (int)ans.size() << '\n';
  for (int& x : ans) cout << x << ' ';
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << ' ' << H;
  debug_out(T...);
}
const int N = 1e6 + 5;
vector<pair<int, int>> v[N];
int vis[N], las[N], dp[N], b[N], tot;
pair<int, int> tag[N * 4], val[N * 4];
void update(int l, int r, int rt, int a, int b, pair<int, int> p) {
  if (l >= a && r <= b) {
    if (tag[rt] < p) tag[rt] = p;
    val[rt] = max(val[rt], p);
    return;
  }
  int m = l + r >> 1;
  if (a <= m) update(l, m, rt << 1, a, b, p);
  if (b > m) update(m + 1, r, rt << 1 | 1, a, b, p);
  val[rt] = max(val[rt << 1], val[rt << 1 | 1]);
}
pair<int, int> query(int l, int r, int rt, int a, int b) {
  if (l >= a && r <= b) return val[rt];
  int m = l + r >> 1;
  pair<int, int> ans = tag[rt];
  if (a <= m) ans = max(ans, query(l, m, rt << 1, a, b));
  if (b > m) ans = max(ans, query(m + 1, r, rt << 1 | 1, a, b));
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1, j, l, r; i <= m; i++) {
    cin >> j >> l >> r;
    v[j].emplace_back(l, r);
    b[++tot] = l, b[++tot] = r;
  }
  sort(b + 1, b + tot + 1);
  tot = unique(b + 1, b + tot + 1) - b - 1;
  int mx = 0, p = 0;
  for (int i = 1; i <= n; i++) {
    pair<int, int> tmp = {0, 0};
    for (auto& e : v[i]) {
      int l = lower_bound(b + 1, b + tot + 1, e.first) - b;
      int r = lower_bound(b + 1, b + tot + 1, e.second) - b;
      auto pi = query(1, tot, 1, l, r);
      tmp = max(tmp, pi);
    }
    dp[i] = tmp.first + 1, las[i] = tmp.second;
    for (auto& e : v[i]) {
      int l = lower_bound(b + 1, b + tot + 1, e.first) - b;
      int r = lower_bound(b + 1, b + tot + 1, e.second) - b;
      update(1, tot, 1, l, r, {dp[i], i});
    }
    if (mx < dp[i]) {
      mx = dp[i];
      p = i;
    }
  }
  while (p) {
    vis[p] = 1;
    p = las[p];
  }
  vector<int> ans;
  for (int i = 1; i <= n; i++)
    if (!vis[i]) ans.push_back(i);
  cout << (int)ans.size() << '\n';
  for (int& x : ans) cout << x << ' ';
  return 0;
}

```