#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-poller", "inline", \
                     "03")
const long double PI = acos(-1);
const long double eps = 1e-6;
const long double pi = acos((long double)-1);
const int MAX = 1e6 + 5;
const int mod = 1e9 + 7;
pair<int, int> st[4 * MAX], lz[4 * MAX];
pair<int, int> ask(pair<int, int> &a, pair<int, int> &b) {
  return a > b ? a : b;
}
void push(int x, pair<int, int> v) {
  if (x < 4 * MAX) {
    st[x] = ask(st[x], v);
    lz[x] = ask(lz[x], v);
  }
}
void update(int x, int L, int R, int lo, int hi, pair<int, int> v) {
  if (lo > R || hi < L) return;
  if (lo <= L && R <= hi) {
    push(x, v);
    return;
  }
  if (lz[x].first) {
    push(2 * x, lz[x]);
    push(2 * x + 1, lz[x]);
    lz[x] = pair<int, int>(0, 0);
  }
  int mid = (L + R) / 2;
  update(2 * x, L, mid, lo, hi, v);
  update(2 * x + 1, mid + 1, R, lo, hi, v);
  st[x] = ask(st[2 * x], st[2 * x + 1]);
}
pair<int, int> query(int x, int L, int R, int lo, int hi) {
  if (lo > R || hi < L) return pair<int, int>(0, 0);
  if (lo <= L && R <= hi) {
    return st[x];
  }
  if (lz[x].first) {
    push(2 * x, lz[x]);
    push(2 * x + 1, lz[x]);
    lz[x] = pair<int, int>(0, 0);
  }
  int mid = (L + R) / 2;
  pair<int, int> v1 = query(2 * x, L, mid, lo, hi);
  pair<int, int> v2 = query(2 * x + 1, mid + 1, R, lo, hi);
  return ask(v1, v2);
}
int t, n, q, a[MAX], b[MAX], c[MAX];
pair<int, int> sol[MAX];
vector<pair<int, int> > pos[MAX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  vector<int> X;
  for (int i = 1; i <= q; i++) {
    cin >> a[i] >> b[i] >> c[i];
    X.push_back(b[i]);
    X.push_back(c[i]);
  }
  sort((X).begin(), (X).end());
  X.erase(unique((X).begin(), (X).end()), X.end());
  for (int i = 1; i <= q; i++) {
    int x = upper_bound((X).begin(), (X).end(), b[i]) - X.begin();
    int y = upper_bound((X).begin(), (X).end(), c[i]) - X.begin();
    pos[a[i]].push_back(pair<int, int>(x, y));
  }
  int L = X.size();
  for (int i = 1; i <= n; i++) {
    pair<int, int> &best = sol[i];
    for (auto [x, y] : pos[i]) {
      pair<int, int> aux = query(1, 1, L, x, y);
      if (aux > best) best = aux;
    }
    for (auto [x, y] : pos[i])
      update(1, 1, L, x, y, pair<int, int>(best.first + 1, i));
  }
  int P = 1;
  for (int i = 2; i <= n; i++)
    if (sol[i] > sol[P]) P = i;
  vector<int> mark(n + 1);
  int cant = n;
  while (P) {
    mark[P] = 1;
    P = sol[P].second;
    cant--;
  }
  cout << cant << '\n';
  for (int i = 1; i <= n; i++) {
    if (!mark[i]) {
      cout << i << " \n"[--cant == 0];
    }
  }
}

### Prompt

Your challenge is to write a cpp solution to 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;
#pragma GCC optimize("Ofast", "unroll-loops", "omit-frame-poller", "inline", \
                     "03")
const long double PI = acos(-1);
const long double eps = 1e-6;
const long double pi = acos((long double)-1);
const int MAX = 1e6 + 5;
const int mod = 1e9 + 7;
pair<int, int> st[4 * MAX], lz[4 * MAX];
pair<int, int> ask(pair<int, int> &a, pair<int, int> &b) {
  return a > b ? a : b;
}
void push(int x, pair<int, int> v) {
  if (x < 4 * MAX) {
    st[x] = ask(st[x], v);
    lz[x] = ask(lz[x], v);
  }
}
void update(int x, int L, int R, int lo, int hi, pair<int, int> v) {
  if (lo > R || hi < L) return;
  if (lo <= L && R <= hi) {
    push(x, v);
    return;
  }
  if (lz[x].first) {
    push(2 * x, lz[x]);
    push(2 * x + 1, lz[x]);
    lz[x] = pair<int, int>(0, 0);
  }
  int mid = (L + R) / 2;
  update(2 * x, L, mid, lo, hi, v);
  update(2 * x + 1, mid + 1, R, lo, hi, v);
  st[x] = ask(st[2 * x], st[2 * x + 1]);
}
pair<int, int> query(int x, int L, int R, int lo, int hi) {
  if (lo > R || hi < L) return pair<int, int>(0, 0);
  if (lo <= L && R <= hi) {
    return st[x];
  }
  if (lz[x].first) {
    push(2 * x, lz[x]);
    push(2 * x + 1, lz[x]);
    lz[x] = pair<int, int>(0, 0);
  }
  int mid = (L + R) / 2;
  pair<int, int> v1 = query(2 * x, L, mid, lo, hi);
  pair<int, int> v2 = query(2 * x + 1, mid + 1, R, lo, hi);
  return ask(v1, v2);
}
int t, n, q, a[MAX], b[MAX], c[MAX];
pair<int, int> sol[MAX];
vector<pair<int, int> > pos[MAX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> q;
  vector<int> X;
  for (int i = 1; i <= q; i++) {
    cin >> a[i] >> b[i] >> c[i];
    X.push_back(b[i]);
    X.push_back(c[i]);
  }
  sort((X).begin(), (X).end());
  X.erase(unique((X).begin(), (X).end()), X.end());
  for (int i = 1; i <= q; i++) {
    int x = upper_bound((X).begin(), (X).end(), b[i]) - X.begin();
    int y = upper_bound((X).begin(), (X).end(), c[i]) - X.begin();
    pos[a[i]].push_back(pair<int, int>(x, y));
  }
  int L = X.size();
  for (int i = 1; i <= n; i++) {
    pair<int, int> &best = sol[i];
    for (auto [x, y] : pos[i]) {
      pair<int, int> aux = query(1, 1, L, x, y);
      if (aux > best) best = aux;
    }
    for (auto [x, y] : pos[i])
      update(1, 1, L, x, y, pair<int, int>(best.first + 1, i));
  }
  int P = 1;
  for (int i = 2; i <= n; i++)
    if (sol[i] > sol[P]) P = i;
  vector<int> mark(n + 1);
  int cant = n;
  while (P) {
    mark[P] = 1;
    P = sol[P].second;
    cant--;
  }
  cout << cant << '\n';
  for (int i = 1; i <= n; i++) {
    if (!mark[i]) {
      cout << i << " \n"[--cant == 0];
    }
  }
}

```