#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e5 + 5;
const int modu = 1e9 + 7;
const double eps = 1e-9;
int n, m;
vector<int> d;
vector<pair<int, int> > seg[maxn];
vector<pair<int, int> > maxt, mint, dp;
vector<bool> tg;
void pushdown(int o) {
  if (tg[o]) {
    if (maxt[(o << 1)] > maxt[o]) {
      maxt[(o << 1)] = maxt[o];
      mint[(o << 1)] = min(mint[(o << 1)], maxt[o]);
      tg[(o << 1)] = 1;
    }
    if (maxt[((o << 1) | 1)] > maxt[o]) {
      maxt[((o << 1) | 1)] = maxt[o];
      mint[((o << 1) | 1)] = min(mint[((o << 1) | 1)], maxt[o]);
      tg[((o << 1) | 1)] = 1;
    }
    tg[o] = 0;
  }
}
void upd(int o, int L, int R, int ql, int qr, pair<int, int> del) {
  if (maxt[o] <= del) return;
  if (ql <= L && R <= qr) {
    maxt[o] = del;
    mint[o] = min(mint[o], del);
    tg[o] = true;
    return;
  }
  pushdown(o);
  int mid = L + R >> 1;
  if (ql <= mid) upd((o << 1), L, mid, ql, qr, del);
  if (qr > mid) upd(((o << 1) | 1), mid + 1, R, ql, qr, del);
  maxt[o] = max(maxt[(o << 1)], maxt[((o << 1) | 1)]);
  mint[o] = min(mint[(o << 1)], mint[((o << 1) | 1)]);
}
pair<int, int> qry(int o, int L, int R, int ql, int qr) {
  if (ql <= L && R <= qr) return mint[o];
  pushdown(o);
  pair<int, int> res = make_pair(1e9, 0);
  int mid = L + R >> 1;
  if (ql <= mid) res = min(res, qry((o << 1), L, mid, ql, qr));
  if (qr > mid) res = min(res, qry(((o << 1) | 1), mid + 1, R, ql, qr));
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  dp.resize(n + 1, make_pair(0, 0));
  for (int i = 0; i < m; ++i) {
    int x, l, r;
    cin >> x >> l >> r;
    seg[x].push_back(make_pair(l, r));
    d.push_back(l);
    d.push_back(r);
  }
  sort(d.begin(), d.end());
  d.erase(unique(d.begin(), d.end()), d.end());
  maxt.resize(8 * (int)d.size(), make_pair(0, 0));
  mint.resize(8 * (int)d.size(), make_pair(0, 0));
  tg.resize(8 * (int)d.size(), 0);
  int tot = (int)d.size();
  for (int i = 1; i <= n; ++i) {
    pair<int, int> res = make_pair(1e9, 0);
    for (auto qj : seg[i]) {
      int l = lower_bound(d.begin(), d.end(), qj.first) - d.begin() + 1;
      int r = lower_bound(d.begin(), d.end(), qj.second) - d.begin() + 1;
      res = min(res, qry(1, 1, tot, l, r));
    }
    res.first += i - 1;
    dp[i] = res;
    res.first -= i;
    res.second = i;
    for (auto qj : seg[i]) {
      int l = lower_bound(d.begin(), d.end(), qj.first) - d.begin() + 1;
      int r = lower_bound(d.begin(), d.end(), qj.second) - d.begin() + 1;
      upd(1, 1, tot, l, r, res);
    }
  }
  int ans = n;
  for (int i = n - 1; i >= 1; --i)
    if (dp[ans].first + n - ans > dp[i].first + n - i) ans = i;
  cout << dp[ans].first + n - ans << "\n";
  for (int i = n; i > ans; --i) cout << i << " ";
  while (ans) {
    for (int i = ans - 1; i > dp[ans].second; --i) cout << i << " ";
    ans = dp[ans].second;
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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 maxn = 5e5 + 5;
const int modu = 1e9 + 7;
const double eps = 1e-9;
int n, m;
vector<int> d;
vector<pair<int, int> > seg[maxn];
vector<pair<int, int> > maxt, mint, dp;
vector<bool> tg;
void pushdown(int o) {
  if (tg[o]) {
    if (maxt[(o << 1)] > maxt[o]) {
      maxt[(o << 1)] = maxt[o];
      mint[(o << 1)] = min(mint[(o << 1)], maxt[o]);
      tg[(o << 1)] = 1;
    }
    if (maxt[((o << 1) | 1)] > maxt[o]) {
      maxt[((o << 1) | 1)] = maxt[o];
      mint[((o << 1) | 1)] = min(mint[((o << 1) | 1)], maxt[o]);
      tg[((o << 1) | 1)] = 1;
    }
    tg[o] = 0;
  }
}
void upd(int o, int L, int R, int ql, int qr, pair<int, int> del) {
  if (maxt[o] <= del) return;
  if (ql <= L && R <= qr) {
    maxt[o] = del;
    mint[o] = min(mint[o], del);
    tg[o] = true;
    return;
  }
  pushdown(o);
  int mid = L + R >> 1;
  if (ql <= mid) upd((o << 1), L, mid, ql, qr, del);
  if (qr > mid) upd(((o << 1) | 1), mid + 1, R, ql, qr, del);
  maxt[o] = max(maxt[(o << 1)], maxt[((o << 1) | 1)]);
  mint[o] = min(mint[(o << 1)], mint[((o << 1) | 1)]);
}
pair<int, int> qry(int o, int L, int R, int ql, int qr) {
  if (ql <= L && R <= qr) return mint[o];
  pushdown(o);
  pair<int, int> res = make_pair(1e9, 0);
  int mid = L + R >> 1;
  if (ql <= mid) res = min(res, qry((o << 1), L, mid, ql, qr));
  if (qr > mid) res = min(res, qry(((o << 1) | 1), mid + 1, R, ql, qr));
  return res;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  dp.resize(n + 1, make_pair(0, 0));
  for (int i = 0; i < m; ++i) {
    int x, l, r;
    cin >> x >> l >> r;
    seg[x].push_back(make_pair(l, r));
    d.push_back(l);
    d.push_back(r);
  }
  sort(d.begin(), d.end());
  d.erase(unique(d.begin(), d.end()), d.end());
  maxt.resize(8 * (int)d.size(), make_pair(0, 0));
  mint.resize(8 * (int)d.size(), make_pair(0, 0));
  tg.resize(8 * (int)d.size(), 0);
  int tot = (int)d.size();
  for (int i = 1; i <= n; ++i) {
    pair<int, int> res = make_pair(1e9, 0);
    for (auto qj : seg[i]) {
      int l = lower_bound(d.begin(), d.end(), qj.first) - d.begin() + 1;
      int r = lower_bound(d.begin(), d.end(), qj.second) - d.begin() + 1;
      res = min(res, qry(1, 1, tot, l, r));
    }
    res.first += i - 1;
    dp[i] = res;
    res.first -= i;
    res.second = i;
    for (auto qj : seg[i]) {
      int l = lower_bound(d.begin(), d.end(), qj.first) - d.begin() + 1;
      int r = lower_bound(d.begin(), d.end(), qj.second) - d.begin() + 1;
      upd(1, 1, tot, l, r, res);
    }
  }
  int ans = n;
  for (int i = n - 1; i >= 1; --i)
    if (dp[ans].first + n - ans > dp[i].first + n - i) ans = i;
  cout << dp[ans].first + n - ans << "\n";
  for (int i = n; i > ans; --i) cout << i << " ";
  while (ans) {
    for (int i = ans - 1; i > dp[ans].second; --i) cout << i << " ";
    ans = dp[ans].second;
  }
  return 0;
}

```