#include <bits/stdc++.h>
using namespace std;
const int N = 6e5;
const int M = N + 7;
const double EPS = 1e-10;
const int MOD = 1e9 + 7;
const long long INF = 1e18 + 17;
int n, m, dp[M];
vector<pair<int, int> > g[M];
int t[4 * M], add[4 * M];
void push(int v, int tl, int tr) {
  int tm = (tl + tr) >> 1;
  if (add[v] != -1) {
    t[v + v] = add[v];
    add[v + v] = add[v];
    t[v + v + 1] = add[v];
    add[v + v + 1] = add[v];
    add[v] = -1;
  }
}
void upd(int v, int tl, int tr, int l, int r, int x) {
  if (l > r) return;
  if (tl == l && tr == r) {
    t[v] = x;
    add[v] = x;
    return;
  }
  push(v, tl, tr);
  int tm = (tl + tr) >> 1;
  upd(v + v, tl, tm, l, min(r, tm), x);
  upd(v + v + 1, tm + 1, tr, max(l, tm + 1), r, x);
  t[v] = max(t[v + v], t[v + v + 1]);
}
int get(int v, int tl, int tr, int l, int r) {
  if (l > r) return 0;
  if (tl == l && tr == r) return t[v];
  push(v, tl, tr);
  int tm = (tl + tr) >> 1;
  return max(get(v + v, tl, tm, l, min(tm, r)),
             get(v + v + 1, tm + 1, tr, max(tm + 1, l), r));
}
void solve() {
  cin >> n >> m;
  map<int, int> com;
  for (int i = 0; i < m; i++) {
    int id, l, r;
    cin >> id >> l >> r;
    com[l]++, com[r]++;
    g[id].push_back({l, r});
  }
  int val = 1;
  for (auto &it : com) it.second = val++;
  for (int i = 1; i <= n; i++)
    for (auto &it : g[i]) it.first = com[it.first], it.second = com[it.second];
  int nn = val - 1;
  dp[1] = 1;
  for (auto iv : g[1]) upd(1, 1, nn, iv.first, iv.second, 1);
  for (int i = 2; i <= n; i++) {
    dp[i] = 1;
    for (auto iv : g[i])
      dp[i] = max(dp[i], get(1, 1, nn, iv.first, iv.second) + 1);
    for (auto iv : g[i]) upd(1, 1, nn, iv.first, iv.second, dp[i]);
  }
  int ans = *max_element(dp + 1, dp + n + 1), last = -1;
  cout << n - ans << '\n';
  for (int i = n; i >= 1; i--) {
    bool ok = 0;
    if (last == -1)
      ok = 1;
    else {
      for (auto iv : g[i])
        for (auto lastv : g[last])
          if (max(iv.first, lastv.first) <= min(iv.second, lastv.second))
            ok = 1;
    }
    if (!ok || dp[i] != ans)
      cout << i << ' ';
    else
      last = i, ans--;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tt = 1;
  while (tt--) {
    solve();
    cout << '\n';
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 = 6e5;
const int M = N + 7;
const double EPS = 1e-10;
const int MOD = 1e9 + 7;
const long long INF = 1e18 + 17;
int n, m, dp[M];
vector<pair<int, int> > g[M];
int t[4 * M], add[4 * M];
void push(int v, int tl, int tr) {
  int tm = (tl + tr) >> 1;
  if (add[v] != -1) {
    t[v + v] = add[v];
    add[v + v] = add[v];
    t[v + v + 1] = add[v];
    add[v + v + 1] = add[v];
    add[v] = -1;
  }
}
void upd(int v, int tl, int tr, int l, int r, int x) {
  if (l > r) return;
  if (tl == l && tr == r) {
    t[v] = x;
    add[v] = x;
    return;
  }
  push(v, tl, tr);
  int tm = (tl + tr) >> 1;
  upd(v + v, tl, tm, l, min(r, tm), x);
  upd(v + v + 1, tm + 1, tr, max(l, tm + 1), r, x);
  t[v] = max(t[v + v], t[v + v + 1]);
}
int get(int v, int tl, int tr, int l, int r) {
  if (l > r) return 0;
  if (tl == l && tr == r) return t[v];
  push(v, tl, tr);
  int tm = (tl + tr) >> 1;
  return max(get(v + v, tl, tm, l, min(tm, r)),
             get(v + v + 1, tm + 1, tr, max(tm + 1, l), r));
}
void solve() {
  cin >> n >> m;
  map<int, int> com;
  for (int i = 0; i < m; i++) {
    int id, l, r;
    cin >> id >> l >> r;
    com[l]++, com[r]++;
    g[id].push_back({l, r});
  }
  int val = 1;
  for (auto &it : com) it.second = val++;
  for (int i = 1; i <= n; i++)
    for (auto &it : g[i]) it.first = com[it.first], it.second = com[it.second];
  int nn = val - 1;
  dp[1] = 1;
  for (auto iv : g[1]) upd(1, 1, nn, iv.first, iv.second, 1);
  for (int i = 2; i <= n; i++) {
    dp[i] = 1;
    for (auto iv : g[i])
      dp[i] = max(dp[i], get(1, 1, nn, iv.first, iv.second) + 1);
    for (auto iv : g[i]) upd(1, 1, nn, iv.first, iv.second, dp[i]);
  }
  int ans = *max_element(dp + 1, dp + n + 1), last = -1;
  cout << n - ans << '\n';
  for (int i = n; i >= 1; i--) {
    bool ok = 0;
    if (last == -1)
      ok = 1;
    else {
      for (auto iv : g[i])
        for (auto lastv : g[last])
          if (max(iv.first, lastv.first) <= min(iv.second, lastv.second))
            ok = 1;
    }
    if (!ok || dp[i] != ans)
      cout << i << ' ';
    else
      last = i, ans--;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int tt = 1;
  while (tt--) {
    solve();
    cout << '\n';
  }
  return 0;
}

```