#include <bits/stdc++.h>
using namespace std;
template <typename T>
void read(vector<T>& a) {
  for (auto& e : a) cin >> e;
}
template <typename T>
void print(vector<T>& a) {
  for (auto& e : a) cout << e + 1 << ' ';
  cout << '\n';
}
vector<pair<int, int> > t, pusher;
void push(int v, int l, int r) {
  if (v >= t.size()) cout << "push: " << v << ' ' << l << ' ' << r << endl;
  if (pusher[v].second == -1) {
    return;
  }
  t[v] = max(t[v], pusher[v]);
  if (l != r) {
    pusher[v * 2] = max(pusher[v * 2], pusher[v]);
    pusher[v * 2 + 1] = max(pusher[v * 2 + 1], pusher[v]);
  }
  pusher[v] = {0, -1};
}
void Set(int v, int l, int r, int ql, int qr, pair<int, int> x) {
  if (v >= t.size()) cout << "set: " << v << ' ' << l << ' ' << r << endl;
  if (qr < l || r < ql) {
    return;
  }
  if (ql <= l && r <= qr) {
    pusher[v] = max(pusher[v], x);
    push(v, l, r);
    return;
  }
  int m = (l + r) / 2;
  Set(v * 2, l, m, ql, qr, x);
  Set(v * 2 + 1, m + 1, r, ql, qr, x);
  t[v] = max(t[v * 2], t[v * 2 + 1]);
}
pair<int, int> get(int v, int l, int r, int ql, int qr) {
  if (v >= t.size()) cout << "get: " << v << ' ' << l << ' ' << r << endl;
  push(v, l, r);
  if (qr < l || r < ql) {
    return {0, -1};
  }
  if (ql <= l && r <= qr) {
    return t[v];
  }
  int m = (l + r) / 2;
  return max(get(v * 2, l, m, ql, qr), get(v * 2 + 1, m + 1, r, ql, qr));
}
void solve() {
  int n, m;
  cin >> n >> m;
  map<int, vector<pair<int, int> > > lev;
  set<int> c;
  for (int i = 0; i < m; ++i) {
    int ll, l, r;
    cin >> ll >> l >> r;
    --ll;
    --l;
    --r;
    lev[ll].push_back({l, r});
    c.insert(l);
    c.insert(r);
  }
  map<int, int> com;
  auto it = c.begin();
  for (int i = 0; i < c.size(); ++i) {
    com[*(it++)] = i;
  }
  for (auto& el : lev) {
    for (auto& s : el.second) {
      s = {com[s.first], com[s.second]};
    }
  }
  int s = 4 * c.size();
  t.resize(s, {0, -1});
  pusher.resize(s, {0, -1});
  vector<int> prev(n + 1, -1);
  for (auto& el : lev) {
    pair<int, int> mx = {0, -1};
    for (auto& seg : el.second) {
      mx = max(mx, get(1, 0, c.size() - 1, seg.first, seg.second));
    }
    mx.first++;
    prev[el.first] = mx.second;
    mx.second = el.first;
    for (auto& seg : el.second) {
      Set(1, 0, c.size() - 1, seg.first, seg.second, mx);
    }
  }
  int curr = get(1, 0, c.size() - 1, 0, c.size() - 1).second;
  vector<bool> ans(n, 1);
  int cnt = n;
  while (curr != -1) {
    ans[curr] = 0;
    cnt--;
    curr = prev[curr];
  }
  cout << cnt << '\n';
  for (int i = 0; i < n; ++i) {
    if (ans[i]) {
      cout << i + 1 << ' ';
    }
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

### Prompt

Please formulate 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;
template <typename T>
void read(vector<T>& a) {
  for (auto& e : a) cin >> e;
}
template <typename T>
void print(vector<T>& a) {
  for (auto& e : a) cout << e + 1 << ' ';
  cout << '\n';
}
vector<pair<int, int> > t, pusher;
void push(int v, int l, int r) {
  if (v >= t.size()) cout << "push: " << v << ' ' << l << ' ' << r << endl;
  if (pusher[v].second == -1) {
    return;
  }
  t[v] = max(t[v], pusher[v]);
  if (l != r) {
    pusher[v * 2] = max(pusher[v * 2], pusher[v]);
    pusher[v * 2 + 1] = max(pusher[v * 2 + 1], pusher[v]);
  }
  pusher[v] = {0, -1};
}
void Set(int v, int l, int r, int ql, int qr, pair<int, int> x) {
  if (v >= t.size()) cout << "set: " << v << ' ' << l << ' ' << r << endl;
  if (qr < l || r < ql) {
    return;
  }
  if (ql <= l && r <= qr) {
    pusher[v] = max(pusher[v], x);
    push(v, l, r);
    return;
  }
  int m = (l + r) / 2;
  Set(v * 2, l, m, ql, qr, x);
  Set(v * 2 + 1, m + 1, r, ql, qr, x);
  t[v] = max(t[v * 2], t[v * 2 + 1]);
}
pair<int, int> get(int v, int l, int r, int ql, int qr) {
  if (v >= t.size()) cout << "get: " << v << ' ' << l << ' ' << r << endl;
  push(v, l, r);
  if (qr < l || r < ql) {
    return {0, -1};
  }
  if (ql <= l && r <= qr) {
    return t[v];
  }
  int m = (l + r) / 2;
  return max(get(v * 2, l, m, ql, qr), get(v * 2 + 1, m + 1, r, ql, qr));
}
void solve() {
  int n, m;
  cin >> n >> m;
  map<int, vector<pair<int, int> > > lev;
  set<int> c;
  for (int i = 0; i < m; ++i) {
    int ll, l, r;
    cin >> ll >> l >> r;
    --ll;
    --l;
    --r;
    lev[ll].push_back({l, r});
    c.insert(l);
    c.insert(r);
  }
  map<int, int> com;
  auto it = c.begin();
  for (int i = 0; i < c.size(); ++i) {
    com[*(it++)] = i;
  }
  for (auto& el : lev) {
    for (auto& s : el.second) {
      s = {com[s.first], com[s.second]};
    }
  }
  int s = 4 * c.size();
  t.resize(s, {0, -1});
  pusher.resize(s, {0, -1});
  vector<int> prev(n + 1, -1);
  for (auto& el : lev) {
    pair<int, int> mx = {0, -1};
    for (auto& seg : el.second) {
      mx = max(mx, get(1, 0, c.size() - 1, seg.first, seg.second));
    }
    mx.first++;
    prev[el.first] = mx.second;
    mx.second = el.first;
    for (auto& seg : el.second) {
      Set(1, 0, c.size() - 1, seg.first, seg.second, mx);
    }
  }
  int curr = get(1, 0, c.size() - 1, 0, c.size() - 1).second;
  vector<bool> ans(n, 1);
  int cnt = n;
  while (curr != -1) {
    ans[curr] = 0;
    cnt--;
    curr = prev[curr];
  }
  cout << cnt << '\n';
  for (int i = 0; i < n; ++i) {
    if (ans[i]) {
      cout << i + 1 << ' ';
    }
  }
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
}

```