#include <bits/stdc++.h>
using namespace std;
const int NMAX = 1000001;
map<int, int> used;
vector<vector<pair<int, int> > > seg(NMAX);
pair<int, int> t[8 * NMAX];
pair<int, int> lazy[8 * NMAX];
void push(int v) {
  if (lazy[v].first == 1e9 + 1) return;
  t[v * 2] = lazy[v];
  t[2 * v + 1] = lazy[v];
  lazy[2 * v] = lazy[v];
  lazy[2 * v + 1] = lazy[v];
  lazy[v] = {1e9 + 1, 1e9 + 1};
}
pair<int, int> get(int v, int l, int r, int st, int fin) {
  if (l > fin || r < st) return {-1e9 + 1, -1e9 + 1};
  if (l >= st && r <= fin) {
    return t[v];
  }
  push(v);
  int m = (l + r) / 2;
  pair<int, int> x = get(2 * v, l, m, st, fin);
  pair<int, int> y = get(2 * v + 1, m + 1, r, st, fin);
  return max(x, y);
}
void update(int v, int l, int r, int st, int fin, int nw, int idx) {
  if (l > fin || st > r) return;
  if (l >= st && r <= fin) {
    t[v] = {nw, idx};
    lazy[v] = {nw, idx};
    return;
  }
  push(v);
  int m = (l + r) / 2;
  update(2 * v, l, m, st, fin, nw, idx);
  update(2 * v + 1, m + 1, r, st, fin, nw, idx);
  t[v] = max(t[2 * v], t[2 * v + 1]);
}
map<int, int> mp;
int R = 2 * 1e6;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> vec;
  for (int i = 1; i <= m; i++) {
    int pos, l, r;
    cin >> pos >> l >> r;
    seg[pos].emplace_back(l, r);
    vec.push_back(l);
    vec.push_back(r);
  }
  sort(vec.begin(), vec.end());
  for (int i = 0; i < vec.size(); i++) {
    mp[vec[i]] = i + 1;
  }
  int dp[n + 1], par[n + 1];
  dp[0] = 0;
  update(1, 1, R, 1, R, 0, 0);
  for (int i = 1; i <= n; i++) {
    pair<int, int> ans = {-1e9 + 1, -1e9 + 1};
    for (int j = 0; j < seg[i].size(); j++) {
      int l = mp[seg[i][j].first], r = mp[seg[i][j].second];
      pair<int, int> x = get(1, 1, R, l, r);
      ans = max(ans, x);
    }
    dp[i] = ans.first + 1;
    par[i] = ans.second;
    for (int j = 0; j < seg[i].size(); j++) {
      int l = mp[seg[i][j].first], r = mp[seg[i][j].second];
      update(1, 1, R, l, r, dp[i], i);
    }
  }
  int mx = -1;
  int mxi;
  for (int i = 1; i <= n; i++) {
    if (mx < dp[i]) {
      mx = dp[i];
      mxi = i;
    }
  }
  set<int> second;
  for (int i = 1; i <= n; i++) second.insert(i);
  cout << n - mx << "\n";
  while (mxi) {
    second.erase(mxi);
    mxi = par[mxi];
  }
  for (int to : second) cout << to << ' ';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int test = 1;
  while (test--) {
    solve();
  }
}

### 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 NMAX = 1000001;
map<int, int> used;
vector<vector<pair<int, int> > > seg(NMAX);
pair<int, int> t[8 * NMAX];
pair<int, int> lazy[8 * NMAX];
void push(int v) {
  if (lazy[v].first == 1e9 + 1) return;
  t[v * 2] = lazy[v];
  t[2 * v + 1] = lazy[v];
  lazy[2 * v] = lazy[v];
  lazy[2 * v + 1] = lazy[v];
  lazy[v] = {1e9 + 1, 1e9 + 1};
}
pair<int, int> get(int v, int l, int r, int st, int fin) {
  if (l > fin || r < st) return {-1e9 + 1, -1e9 + 1};
  if (l >= st && r <= fin) {
    return t[v];
  }
  push(v);
  int m = (l + r) / 2;
  pair<int, int> x = get(2 * v, l, m, st, fin);
  pair<int, int> y = get(2 * v + 1, m + 1, r, st, fin);
  return max(x, y);
}
void update(int v, int l, int r, int st, int fin, int nw, int idx) {
  if (l > fin || st > r) return;
  if (l >= st && r <= fin) {
    t[v] = {nw, idx};
    lazy[v] = {nw, idx};
    return;
  }
  push(v);
  int m = (l + r) / 2;
  update(2 * v, l, m, st, fin, nw, idx);
  update(2 * v + 1, m + 1, r, st, fin, nw, idx);
  t[v] = max(t[2 * v], t[2 * v + 1]);
}
map<int, int> mp;
int R = 2 * 1e6;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<int> vec;
  for (int i = 1; i <= m; i++) {
    int pos, l, r;
    cin >> pos >> l >> r;
    seg[pos].emplace_back(l, r);
    vec.push_back(l);
    vec.push_back(r);
  }
  sort(vec.begin(), vec.end());
  for (int i = 0; i < vec.size(); i++) {
    mp[vec[i]] = i + 1;
  }
  int dp[n + 1], par[n + 1];
  dp[0] = 0;
  update(1, 1, R, 1, R, 0, 0);
  for (int i = 1; i <= n; i++) {
    pair<int, int> ans = {-1e9 + 1, -1e9 + 1};
    for (int j = 0; j < seg[i].size(); j++) {
      int l = mp[seg[i][j].first], r = mp[seg[i][j].second];
      pair<int, int> x = get(1, 1, R, l, r);
      ans = max(ans, x);
    }
    dp[i] = ans.first + 1;
    par[i] = ans.second;
    for (int j = 0; j < seg[i].size(); j++) {
      int l = mp[seg[i][j].first], r = mp[seg[i][j].second];
      update(1, 1, R, l, r, dp[i], i);
    }
  }
  int mx = -1;
  int mxi;
  for (int i = 1; i <= n; i++) {
    if (mx < dp[i]) {
      mx = dp[i];
      mxi = i;
    }
  }
  set<int> second;
  for (int i = 1; i <= n; i++) second.insert(i);
  cout << n - mx << "\n";
  while (mxi) {
    second.erase(mxi);
    mxi = par[mxi];
  }
  for (int to : second) cout << to << ' ';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int test = 1;
  while (test--) {
    solve();
  }
}

```