#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 10;
const int M = 1e9 + 7;
int n, m, ind, l, r;
int prv[N], a[2 * N];
struct Interv {
  int ind, l, r;
} interv[N];
pair<int, int> arb[8 * N], lazy[8 * N];
bitset<N> mark;
void push(int node, int l, int r) {
  if (l == r || !lazy[node].first) return;
  lazy[node * 2] = arb[node * 2] = lazy[node];
  lazy[node * 2 + 1] = arb[node * 2 + 1] = lazy[node];
  lazy[node] = {0, 0};
}
void setValue(int node, int l, int r, int L, int R, pair<int, int> value) {
  if (L > R) return;
  if (L <= l && r <= R) {
    arb[node] = lazy[node] = value;
    return;
  }
  int mid = (l + r) / 2;
  push(node, l, r);
  setValue(node * 2, l, mid, L, min(R, mid), value);
  setValue(node * 2 + 1, mid + 1, r, max(L, mid + 1), R, value);
  arb[node] = max(arb[node * 2], arb[node * 2 + 1]);
}
pair<int, int> query(int node, int l, int r, int L, int R) {
  if (L > R) return {-1, 0};
  if (L <= l && r <= R) return arb[node];
  push(node, l, r);
  int mid = (l + r) / 2;
  return max(query(node * 2, l, mid, L, min(R, mid)),
             query(node * 2 + 1, mid + 1, r, max(L, mid + 1), R));
}
bool inline cmp(Interv a, Interv b) { return a.ind < b.ind; }
signed main() {
  int t = 0, i, j, row;
  cin >> n >> m;
  for (i = 1; i <= m; i++) {
    cin >> ind >> l >> r;
    interv[i] = {ind, l, r};
    a[++t] = l;
    a[++t] = r;
  }
  sort(a + 1, a + t + 1);
  int y = 0;
  for (i = 2; i <= t + 1; i++)
    if (a[i] != a[i - 1]) a[++y] = a[i - 1];
  t = y;
  for (i = 1; i <= m; i++) {
    interv[i].l = lower_bound(a + 1, a + t + 1, interv[i].l) - a;
    interv[i].r = lower_bound(a + 1, a + t + 1, interv[i].r) - a;
  }
  sort(interv + 1, interv + m + 1, cmp);
  pair<int, int> ans = {0, 0};
  pair<int, int> mx = {0, 0};
  for (row = 1; row <= m + 1; row++) {
    if (interv[row].ind != interv[row - 1].ind && row > 1) {
      for (i = row - 1; interv[i].ind == interv[row - 1].ind; i--)
        setValue(1, 1, 2 * m, interv[i].l, interv[i].r,
                 {mx.first + 1, interv[i].ind});
      prv[interv[row - 1].ind] = mx.second;
      ans = max(ans, {mx.first + 1, interv[row - 1].ind});
      mx = {0, 0};
    }
    if (row <= m)
      mx = max(mx, query(1, 1, 2 * m, interv[row].l, interv[row].r));
  }
  int choose = 0;
  row = ans.second;
  do {
    ++choose;
    mark[row] = 1;
    row = prv[row];
  } while (row);
  cout << n - choose << '\n';
  for (i = 1; i <= n; i++)
    if (!mark[i]) cout << i << ' ';
  return 0;
}

### Prompt

Generate 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;
const int N = 3e5 + 10;
const int M = 1e9 + 7;
int n, m, ind, l, r;
int prv[N], a[2 * N];
struct Interv {
  int ind, l, r;
} interv[N];
pair<int, int> arb[8 * N], lazy[8 * N];
bitset<N> mark;
void push(int node, int l, int r) {
  if (l == r || !lazy[node].first) return;
  lazy[node * 2] = arb[node * 2] = lazy[node];
  lazy[node * 2 + 1] = arb[node * 2 + 1] = lazy[node];
  lazy[node] = {0, 0};
}
void setValue(int node, int l, int r, int L, int R, pair<int, int> value) {
  if (L > R) return;
  if (L <= l && r <= R) {
    arb[node] = lazy[node] = value;
    return;
  }
  int mid = (l + r) / 2;
  push(node, l, r);
  setValue(node * 2, l, mid, L, min(R, mid), value);
  setValue(node * 2 + 1, mid + 1, r, max(L, mid + 1), R, value);
  arb[node] = max(arb[node * 2], arb[node * 2 + 1]);
}
pair<int, int> query(int node, int l, int r, int L, int R) {
  if (L > R) return {-1, 0};
  if (L <= l && r <= R) return arb[node];
  push(node, l, r);
  int mid = (l + r) / 2;
  return max(query(node * 2, l, mid, L, min(R, mid)),
             query(node * 2 + 1, mid + 1, r, max(L, mid + 1), R));
}
bool inline cmp(Interv a, Interv b) { return a.ind < b.ind; }
signed main() {
  int t = 0, i, j, row;
  cin >> n >> m;
  for (i = 1; i <= m; i++) {
    cin >> ind >> l >> r;
    interv[i] = {ind, l, r};
    a[++t] = l;
    a[++t] = r;
  }
  sort(a + 1, a + t + 1);
  int y = 0;
  for (i = 2; i <= t + 1; i++)
    if (a[i] != a[i - 1]) a[++y] = a[i - 1];
  t = y;
  for (i = 1; i <= m; i++) {
    interv[i].l = lower_bound(a + 1, a + t + 1, interv[i].l) - a;
    interv[i].r = lower_bound(a + 1, a + t + 1, interv[i].r) - a;
  }
  sort(interv + 1, interv + m + 1, cmp);
  pair<int, int> ans = {0, 0};
  pair<int, int> mx = {0, 0};
  for (row = 1; row <= m + 1; row++) {
    if (interv[row].ind != interv[row - 1].ind && row > 1) {
      for (i = row - 1; interv[i].ind == interv[row - 1].ind; i--)
        setValue(1, 1, 2 * m, interv[i].l, interv[i].r,
                 {mx.first + 1, interv[i].ind});
      prv[interv[row - 1].ind] = mx.second;
      ans = max(ans, {mx.first + 1, interv[row - 1].ind});
      mx = {0, 0};
    }
    if (row <= m)
      mx = max(mx, query(1, 1, 2 * m, interv[row].l, interv[row].r));
  }
  int choose = 0;
  row = ans.second;
  do {
    ++choose;
    mark[row] = 1;
    row = prv[row];
  } while (row);
  cout << n - choose << '\n';
  for (i = 1; i <= n; i++)
    if (!mark[i]) cout << i << ' ';
  return 0;
}

```