#include <bits/stdc++.h>
using namespace std;
const int N = 600005;
vector<array<int, 2>> tree(4 * N, {0, 0});
vector<array<int, 2>> lazy(4 * N, {0, 0});
void apply(int idx, array<int, 2> a) {
  tree[idx] = max(tree[idx], a);
  lazy[idx] = max(lazy[idx], a);
}
void push(int idx) {
  for (int i = 0; i < 2; i++) apply(2 * idx + i, lazy[idx]);
}
void update(int idx, int l, int r, int ql, int qr, array<int, 2> a) {
  if (ql > qr) return;
  if (l == ql && r == qr)
    apply(idx, a);
  else {
    int m = (l + r) / 2;
    push(idx);
    update(2 * idx, l, m, ql, min(qr, m), a);
    update(2 * idx + 1, m + 1, r, max(ql, m + 1), qr, a);
    tree[idx] = max(tree[2 * idx], tree[2 * idx + 1]);
  }
}
array<int, 2> query(int idx, int l, int r, int ql, int qr) {
  if (ql > qr) return {0, 0};
  if (l == ql && r == qr) return tree[idx];
  int m = (l + r) / 2;
  push(idx);
  return max(query(2 * idx, l, m, ql, min(qr, m)),
             query(2 * idx + 1, m + 1, r, max(ql, m + 1), qr));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<array<int, 2>> v[n + 1];
  vector<array<int, 4>> h;
  for (int i = 0; i < m; i++) {
    int x, l, r;
    cin >> x >> l >> r;
    h.push_back({l, x, (int)v[x].size(), 0});
    h.push_back({r, x, (int)v[x].size(), 1});
    v[x].push_back({l, r});
  }
  sort(h.begin(), h.end());
  int id = 1;
  for (int i = 0; i < 2 * m; i++) {
    if (i > 0 && h[i][0] != h[i - 1][0]) id++;
    v[h[i][1]][h[i][2]][h[i][3]] = id;
  }
  vector<array<int, 2>> best(n + 1, {0, 0});
  array<int, 2> opt = {0, 0};
  for (int i = 1; i <= n; i++) {
    for (auto [l, r] : v[i]) best[i] = max(best[i], query(1, 1, id, l, r));
    best[i][0]++;
    for (auto [l, r] : v[i]) update(1, 1, id, l, r, {best[i][0], i});
    opt = max(opt, {best[i][0], i});
  }
  vector<bool> vis(n + 1, 0);
  int x = opt[1];
  while (x > 0) {
    vis[x] = 1;
    x = best[x][1];
  }
  cout << n - opt[0] << "\n";
  for (int i = 1; i <= n; i++)
    if (vis[i] == 0) cout << i << " ";
  cout << "\n";
  return 0;
}

### Prompt

In cpp, your task is to solve 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 = 600005;
vector<array<int, 2>> tree(4 * N, {0, 0});
vector<array<int, 2>> lazy(4 * N, {0, 0});
void apply(int idx, array<int, 2> a) {
  tree[idx] = max(tree[idx], a);
  lazy[idx] = max(lazy[idx], a);
}
void push(int idx) {
  for (int i = 0; i < 2; i++) apply(2 * idx + i, lazy[idx]);
}
void update(int idx, int l, int r, int ql, int qr, array<int, 2> a) {
  if (ql > qr) return;
  if (l == ql && r == qr)
    apply(idx, a);
  else {
    int m = (l + r) / 2;
    push(idx);
    update(2 * idx, l, m, ql, min(qr, m), a);
    update(2 * idx + 1, m + 1, r, max(ql, m + 1), qr, a);
    tree[idx] = max(tree[2 * idx], tree[2 * idx + 1]);
  }
}
array<int, 2> query(int idx, int l, int r, int ql, int qr) {
  if (ql > qr) return {0, 0};
  if (l == ql && r == qr) return tree[idx];
  int m = (l + r) / 2;
  push(idx);
  return max(query(2 * idx, l, m, ql, min(qr, m)),
             query(2 * idx + 1, m + 1, r, max(ql, m + 1), qr));
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<array<int, 2>> v[n + 1];
  vector<array<int, 4>> h;
  for (int i = 0; i < m; i++) {
    int x, l, r;
    cin >> x >> l >> r;
    h.push_back({l, x, (int)v[x].size(), 0});
    h.push_back({r, x, (int)v[x].size(), 1});
    v[x].push_back({l, r});
  }
  sort(h.begin(), h.end());
  int id = 1;
  for (int i = 0; i < 2 * m; i++) {
    if (i > 0 && h[i][0] != h[i - 1][0]) id++;
    v[h[i][1]][h[i][2]][h[i][3]] = id;
  }
  vector<array<int, 2>> best(n + 1, {0, 0});
  array<int, 2> opt = {0, 0};
  for (int i = 1; i <= n; i++) {
    for (auto [l, r] : v[i]) best[i] = max(best[i], query(1, 1, id, l, r));
    best[i][0]++;
    for (auto [l, r] : v[i]) update(1, 1, id, l, r, {best[i][0], i});
    opt = max(opt, {best[i][0], i});
  }
  vector<bool> vis(n + 1, 0);
  int x = opt[1];
  while (x > 0) {
    vis[x] = 1;
    x = best[x][1];
  }
  cout << n - opt[0] << "\n";
  for (int i = 1; i <= n; i++)
    if (vis[i] == 0) cout << i << " ";
  cout << "\n";
  return 0;
}

```