#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long inf = 1e9 + 7;
const double eps = 1e-9;
struct segtree {
  int n;
  pair<int, int> t[2400005], lz[2400005];
  const pair<int, int> neutral = {0, -1};
  void init(int n) {
    this->n = n;
    fill(t + 1, t + 4 * n + 5, neutral);
    fill(lz + 1, lz + 4 * n + 5, neutral);
  }
  pair<int, int> comb(pair<int, int> u, pair<int, int> v) { return max(u, v); }
  void pull(int u) { t[u] = comb(t[2 * u], t[2 * u + 1]); }
  void push(int u, int tl, int tr) {
    if (lz[u] != neutral) {
      t[u] = max(t[u], lz[u]);
      if (tl != tr) {
        lz[2 * u] = max(lz[2 * u], lz[u]);
        lz[2 * u + 1] = max(lz[2 * u + 1], lz[u]);
      }
      lz[u] = neutral;
    }
  }
  void upd(int u, int tl, int tr, int l, int r, pair<int, int> val) {
    push(u, tl, tr);
    if (l > r)
      return;
    else if (l == tl && r == tr) {
      lz[u] = max(lz[u], val);
      push(u, tl, tr);
    } else {
      int tm = (tl + tr) / 2;
      upd(2 * u, tl, tm, l, min(r, tm), val);
      upd(2 * u + 1, tm + 1, tr, max(l, tm + 1), r, val);
      pull(u);
    }
  }
  void upd(int l, int r, pair<int, int> val) { upd(1, 1, n, l, r, val); }
  pair<int, int> query(int u, int tl, int tr, int l, int r) {
    if (l > r) return neutral;
    push(u, tl, tr);
    if (l == tl && r == tr)
      return t[u];
    else {
      int tm = (tl + tr) / 2;
      return comb(query(2 * u, tl, tm, l, min(r, tm)),
                  query(2 * u + 1, tm + 1, tr, max(l, tm + 1), r));
    }
  }
  pair<int, int> query(int l, int r) { return query(1, 1, n, l, r); }
} ds;
int n, m;
vector<pair<int, int> > r[300005];
vector<int> comp;
int p[300005];
bool vis[300005];
int ans, cur;
inline int compress(int x) {
  return upper_bound(comp.begin(), comp.end(), x) - comp.begin();
}
void solve() {
  cin >> n >> m;
  for (int i = 1, row, le, ri; i <= m; ++i) {
    cin >> row >> le >> ri;
    r[row].push_back({le, ri});
    comp.push_back(le);
    comp.push_back(ri);
  }
  sort(comp.begin(), comp.end());
  comp.resize(unique(comp.begin(), comp.end()) - comp.begin());
  for (int i = 1; i <= n; ++i)
    for (auto &x : r[i]) x = {compress(x.first), compress(x.second)};
  ds.init(comp.size());
  for (int i = 1; i <= n; ++i) {
    pair<int, int> mx = {0, -1};
    for (auto x : r[i]) mx = max(mx, ds.query(x.first, x.second));
    p[i] = mx.second;
    for (auto x : r[i]) ds.upd(x.first, x.second, {mx.first + 1, i});
  }
  tie(ans, cur) = ds.query(1, comp.size());
  memset(vis, 0, sizeof(vis));
  for (; cur != -1; cur = p[cur]) vis[cur] = 1;
  cout << n - ans << "\n";
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) cout << i << " ";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int test = 1;
  for (int tc = 1; tc <= test; ++tc) solve();
  return 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;
const long long mod = 1e9 + 7;
const long long inf = 1e9 + 7;
const double eps = 1e-9;
struct segtree {
  int n;
  pair<int, int> t[2400005], lz[2400005];
  const pair<int, int> neutral = {0, -1};
  void init(int n) {
    this->n = n;
    fill(t + 1, t + 4 * n + 5, neutral);
    fill(lz + 1, lz + 4 * n + 5, neutral);
  }
  pair<int, int> comb(pair<int, int> u, pair<int, int> v) { return max(u, v); }
  void pull(int u) { t[u] = comb(t[2 * u], t[2 * u + 1]); }
  void push(int u, int tl, int tr) {
    if (lz[u] != neutral) {
      t[u] = max(t[u], lz[u]);
      if (tl != tr) {
        lz[2 * u] = max(lz[2 * u], lz[u]);
        lz[2 * u + 1] = max(lz[2 * u + 1], lz[u]);
      }
      lz[u] = neutral;
    }
  }
  void upd(int u, int tl, int tr, int l, int r, pair<int, int> val) {
    push(u, tl, tr);
    if (l > r)
      return;
    else if (l == tl && r == tr) {
      lz[u] = max(lz[u], val);
      push(u, tl, tr);
    } else {
      int tm = (tl + tr) / 2;
      upd(2 * u, tl, tm, l, min(r, tm), val);
      upd(2 * u + 1, tm + 1, tr, max(l, tm + 1), r, val);
      pull(u);
    }
  }
  void upd(int l, int r, pair<int, int> val) { upd(1, 1, n, l, r, val); }
  pair<int, int> query(int u, int tl, int tr, int l, int r) {
    if (l > r) return neutral;
    push(u, tl, tr);
    if (l == tl && r == tr)
      return t[u];
    else {
      int tm = (tl + tr) / 2;
      return comb(query(2 * u, tl, tm, l, min(r, tm)),
                  query(2 * u + 1, tm + 1, tr, max(l, tm + 1), r));
    }
  }
  pair<int, int> query(int l, int r) { return query(1, 1, n, l, r); }
} ds;
int n, m;
vector<pair<int, int> > r[300005];
vector<int> comp;
int p[300005];
bool vis[300005];
int ans, cur;
inline int compress(int x) {
  return upper_bound(comp.begin(), comp.end(), x) - comp.begin();
}
void solve() {
  cin >> n >> m;
  for (int i = 1, row, le, ri; i <= m; ++i) {
    cin >> row >> le >> ri;
    r[row].push_back({le, ri});
    comp.push_back(le);
    comp.push_back(ri);
  }
  sort(comp.begin(), comp.end());
  comp.resize(unique(comp.begin(), comp.end()) - comp.begin());
  for (int i = 1; i <= n; ++i)
    for (auto &x : r[i]) x = {compress(x.first), compress(x.second)};
  ds.init(comp.size());
  for (int i = 1; i <= n; ++i) {
    pair<int, int> mx = {0, -1};
    for (auto x : r[i]) mx = max(mx, ds.query(x.first, x.second));
    p[i] = mx.second;
    for (auto x : r[i]) ds.upd(x.first, x.second, {mx.first + 1, i});
  }
  tie(ans, cur) = ds.query(1, comp.size());
  memset(vis, 0, sizeof(vis));
  for (; cur != -1; cur = p[cur]) vis[cur] = 1;
  cout << n - ans << "\n";
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) cout << i << " ";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int test = 1;
  for (int tc = 1; tc <= test; ++tc) solve();
  return 0;
}

```