#include <bits/stdc++.h>
using namespace std;
const int off = 1 << 20, maxn = 3e5;
int n, m, k, l, r;
vector<int> v;
vector<pair<int, int> > inte[maxn];
pair<int, int> tour[2 * off];
pair<int, int> prop[2 * off];
pair<int, int> dp[maxn];
int bio[maxn];
void f(int x, int v, int ind) {
  if (v > tour[x].first) {
    tour[x] = {v, ind};
  }
  if (v > prop[x].first) {
    prop[x] = {v, ind};
  }
}
void push(int x) {
  if (x >= off || prop[x].first == 0) return;
  int v = prop[x].first;
  int ind = prop[x].second;
  prop[x].first = 0;
  f(2 * x, v, ind);
  f(2 * x + 1, v, ind);
}
void update(int x, int lo, int hi, int l, int r, int v, int ind) {
  push(x);
  if (hi <= l || r <= lo) return;
  if (l <= lo && hi <= r) {
    f(x, v, ind);
    return;
  }
  int mid = (lo + hi) / 2;
  update(x * 2, lo, mid, l, r, v, ind);
  update(x * 2 + 1, mid, hi, l, r, v, ind);
  pair<int, int> t1 = tour[x * 2];
  pair<int, int> t2 = tour[x * 2 + 1];
  if (t1.first > t2.first)
    tour[x] = t1;
  else
    tour[x] = t2;
}
pair<int, int> query(int x, int lo, int hi, int l, int r) {
  push(x);
  if (hi <= l || r <= lo) return {0, -1};
  if (l <= lo && hi <= r) return tour[x];
  int mid = (lo + hi) / 2;
  pair<int, int> q1 = query(x * 2, lo, mid, l, r);
  pair<int, int> q2 = query(x * 2 + 1, mid, hi, l, r);
  if (q1.first > q2.first) return q1;
  return q2;
}
void ispis() {
  for (int i = 1; i < 2 * off; i++) {
    if ((i & (i - 1)) == 0) cout << endl;
    cout << "[" << tour[i].first << ' ' << prop[i].first << ' '
         << tour[i].second << "] ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> k >> l >> r;
    k--;
    inte[k].push_back({l, r});
    v.push_back(l);
    v.push_back(r);
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < inte[i].size(); j++) {
      inte[i][j].first =
          lower_bound(v.begin(), v.end(), inte[i][j].first) - v.begin();
      inte[i][j].second =
          lower_bound(v.begin(), v.end(), inte[i][j].second) - v.begin();
    }
  }
  for (int i = 1; i < 2 * off; i++) tour[i].second = -1;
  for (int i = 0; i < n; i++) {
    dp[i].second = -1;
    for (int j = 0; j < inte[i].size(); j++) {
      pair<int, int> q =
          query(1, 0, off, inte[i][j].first, inte[i][j].second + 1);
      if (q.first > dp[i].first) dp[i] = q;
    }
    dp[i].first++;
    for (int j = 0; j < inte[i].size(); j++) {
      update(1, 0, off, inte[i][j].first, inte[i][j].second + 1, dp[i].first,
             i);
    }
  }
  int x = n;
  for (int i = 0; i < n; i++) {
    if (dp[i].first > dp[x].first) x = i;
  }
  cout << n - dp[x].first << "\n";
  while (x != -1) {
    bio[x] = 1;
    x = dp[x].second;
  }
  for (int i = 0; i < n; i++) {
    if (!bio[i]) cout << i + 1 << ' ';
  }
  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 int off = 1 << 20, maxn = 3e5;
int n, m, k, l, r;
vector<int> v;
vector<pair<int, int> > inte[maxn];
pair<int, int> tour[2 * off];
pair<int, int> prop[2 * off];
pair<int, int> dp[maxn];
int bio[maxn];
void f(int x, int v, int ind) {
  if (v > tour[x].first) {
    tour[x] = {v, ind};
  }
  if (v > prop[x].first) {
    prop[x] = {v, ind};
  }
}
void push(int x) {
  if (x >= off || prop[x].first == 0) return;
  int v = prop[x].first;
  int ind = prop[x].second;
  prop[x].first = 0;
  f(2 * x, v, ind);
  f(2 * x + 1, v, ind);
}
void update(int x, int lo, int hi, int l, int r, int v, int ind) {
  push(x);
  if (hi <= l || r <= lo) return;
  if (l <= lo && hi <= r) {
    f(x, v, ind);
    return;
  }
  int mid = (lo + hi) / 2;
  update(x * 2, lo, mid, l, r, v, ind);
  update(x * 2 + 1, mid, hi, l, r, v, ind);
  pair<int, int> t1 = tour[x * 2];
  pair<int, int> t2 = tour[x * 2 + 1];
  if (t1.first > t2.first)
    tour[x] = t1;
  else
    tour[x] = t2;
}
pair<int, int> query(int x, int lo, int hi, int l, int r) {
  push(x);
  if (hi <= l || r <= lo) return {0, -1};
  if (l <= lo && hi <= r) return tour[x];
  int mid = (lo + hi) / 2;
  pair<int, int> q1 = query(x * 2, lo, mid, l, r);
  pair<int, int> q2 = query(x * 2 + 1, mid, hi, l, r);
  if (q1.first > q2.first) return q1;
  return q2;
}
void ispis() {
  for (int i = 1; i < 2 * off; i++) {
    if ((i & (i - 1)) == 0) cout << endl;
    cout << "[" << tour[i].first << ' ' << prop[i].first << ' '
         << tour[i].second << "] ";
  }
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> k >> l >> r;
    k--;
    inte[k].push_back({l, r});
    v.push_back(l);
    v.push_back(r);
  }
  sort(v.begin(), v.end());
  v.erase(unique(v.begin(), v.end()), v.end());
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < inte[i].size(); j++) {
      inte[i][j].first =
          lower_bound(v.begin(), v.end(), inte[i][j].first) - v.begin();
      inte[i][j].second =
          lower_bound(v.begin(), v.end(), inte[i][j].second) - v.begin();
    }
  }
  for (int i = 1; i < 2 * off; i++) tour[i].second = -1;
  for (int i = 0; i < n; i++) {
    dp[i].second = -1;
    for (int j = 0; j < inte[i].size(); j++) {
      pair<int, int> q =
          query(1, 0, off, inte[i][j].first, inte[i][j].second + 1);
      if (q.first > dp[i].first) dp[i] = q;
    }
    dp[i].first++;
    for (int j = 0; j < inte[i].size(); j++) {
      update(1, 0, off, inte[i][j].first, inte[i][j].second + 1, dp[i].first,
             i);
    }
  }
  int x = n;
  for (int i = 0; i < n; i++) {
    if (dp[i].first > dp[x].first) x = i;
  }
  cout << n - dp[x].first << "\n";
  while (x != -1) {
    bio[x] = 1;
    x = dp[x].second;
  }
  for (int i = 0; i < n; i++) {
    if (!bio[i]) cout << i + 1 << ' ';
  }
  return 0;
}

```