#include <bits/stdc++.h>
using namespace std;
pair<int, int> seg[2100000], lazy[2100000];
pair<int, int> Merge(pair<int, int> l, pair<int, int> r) {
  if (l < r)
    return l;
  else
    return r;
}
void Propagate(int n, int l, int r) {
  if (lazy[n] != make_pair(0x3f3f3f3f, 0)) {
    if (l != r) {
      lazy[n << 1] = Merge(lazy[n << 1], lazy[n]);
      lazy[n << 1 | 1] = Merge(lazy[n << 1 | 1], lazy[n]);
    }
    seg[n] = Merge(seg[n], lazy[n]);
    lazy[n] = make_pair(0x3f3f3f3f, 0);
  }
}
pair<int, int> Update(int L, int R, pair<int, int> val, int n, int l, int r) {
  Propagate(n, l, r);
  if (r < L || R < l) return seg[n];
  if (L <= l && r <= R) {
    lazy[n] = Merge(lazy[n], val);
    Propagate(n, l, r);
    return seg[n];
  }
  int mid = l + r >> 1;
  return seg[n] = Merge(Update(L, R, val, n << 1, l, mid),
                        Update(L, R, val, n << 1 | 1, mid + 1, r));
}
pair<int, int> Query(int L, int R, int n, int l, int r) {
  Propagate(n, l, r);
  if (r < L || R < l) return make_pair(0x3f3f3f3f, 0);
  if (L <= l && r <= R) return seg[n];
  int mid = l + r >> 1;
  return Merge(Query(L, R, n << 1, l, mid),
               Query(L, R, n << 1 | 1, mid + 1, r));
}
int N, M;
vector<pair<int, int> > v[300001];
pair<int, int> dp[300001];
int main() {
  for (int i = 0; i < 2100000; i++) seg[i] = lazy[i] = make_pair(0x3f3f3f3f, 0);
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(ios::badbit | ios::failbit);
  cin >> N >> M;
  vector<int> com;
  for (int i = 1; i <= M; i++) {
    int n, l, r;
    cin >> n >> l >> r;
    r++;
    v[n].push_back({l, r});
    com.push_back(l);
    com.push_back(r);
  }
  sort(com.begin(), com.end());
  com.erase(unique(com.begin(), com.end()), com.end());
  for (int i = 1; i <= N; i++)
    for (auto &[l, r] : v[i]) {
      l = lower_bound(com.begin(), com.end(), l) - com.begin();
      r = lower_bound(com.begin(), com.end(), r) - com.begin();
    }
  for (int i = 1; i <= N; i++) {
    dp[i] = {i - 1, 0};
    for (auto [l, r] : v[i]) {
      auto it = Query(l, r - 1, 1, 0, (int)com.size() - 1);
      dp[i] = min(dp[i], make_pair(it.first + i, it.second));
    }
    for (auto [l, r] : v[i]) {
      Update(l, r - 1, make_pair(dp[i].first - i - 1, i), 1, 0,
             (int)com.size() - 1);
    }
  }
  int res = 0x3f3f3f3f, n = 0;
  for (int i = 1; i <= N; i++) {
    int val = dp[i].first + (N - i);
    if (res > val) res = val, n = i;
  }
  set<int> s;
  for (int i = 1; i <= N; i++) s.insert(i);
  while (n) {
    s.erase(n);
    n = dp[n].second;
  }
  cout << res << "\n";
  if (res) {
    for (int i : s) 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;
pair<int, int> seg[2100000], lazy[2100000];
pair<int, int> Merge(pair<int, int> l, pair<int, int> r) {
  if (l < r)
    return l;
  else
    return r;
}
void Propagate(int n, int l, int r) {
  if (lazy[n] != make_pair(0x3f3f3f3f, 0)) {
    if (l != r) {
      lazy[n << 1] = Merge(lazy[n << 1], lazy[n]);
      lazy[n << 1 | 1] = Merge(lazy[n << 1 | 1], lazy[n]);
    }
    seg[n] = Merge(seg[n], lazy[n]);
    lazy[n] = make_pair(0x3f3f3f3f, 0);
  }
}
pair<int, int> Update(int L, int R, pair<int, int> val, int n, int l, int r) {
  Propagate(n, l, r);
  if (r < L || R < l) return seg[n];
  if (L <= l && r <= R) {
    lazy[n] = Merge(lazy[n], val);
    Propagate(n, l, r);
    return seg[n];
  }
  int mid = l + r >> 1;
  return seg[n] = Merge(Update(L, R, val, n << 1, l, mid),
                        Update(L, R, val, n << 1 | 1, mid + 1, r));
}
pair<int, int> Query(int L, int R, int n, int l, int r) {
  Propagate(n, l, r);
  if (r < L || R < l) return make_pair(0x3f3f3f3f, 0);
  if (L <= l && r <= R) return seg[n];
  int mid = l + r >> 1;
  return Merge(Query(L, R, n << 1, l, mid),
               Query(L, R, n << 1 | 1, mid + 1, r));
}
int N, M;
vector<pair<int, int> > v[300001];
pair<int, int> dp[300001];
int main() {
  for (int i = 0; i < 2100000; i++) seg[i] = lazy[i] = make_pair(0x3f3f3f3f, 0);
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin.exceptions(ios::badbit | ios::failbit);
  cin >> N >> M;
  vector<int> com;
  for (int i = 1; i <= M; i++) {
    int n, l, r;
    cin >> n >> l >> r;
    r++;
    v[n].push_back({l, r});
    com.push_back(l);
    com.push_back(r);
  }
  sort(com.begin(), com.end());
  com.erase(unique(com.begin(), com.end()), com.end());
  for (int i = 1; i <= N; i++)
    for (auto &[l, r] : v[i]) {
      l = lower_bound(com.begin(), com.end(), l) - com.begin();
      r = lower_bound(com.begin(), com.end(), r) - com.begin();
    }
  for (int i = 1; i <= N; i++) {
    dp[i] = {i - 1, 0};
    for (auto [l, r] : v[i]) {
      auto it = Query(l, r - 1, 1, 0, (int)com.size() - 1);
      dp[i] = min(dp[i], make_pair(it.first + i, it.second));
    }
    for (auto [l, r] : v[i]) {
      Update(l, r - 1, make_pair(dp[i].first - i - 1, i), 1, 0,
             (int)com.size() - 1);
    }
  }
  int res = 0x3f3f3f3f, n = 0;
  for (int i = 1; i <= N; i++) {
    int val = dp[i].first + (N - i);
    if (res > val) res = val, n = i;
  }
  set<int> s;
  for (int i = 1; i <= N; i++) s.insert(i);
  while (n) {
    s.erase(n);
    n = dp[n].second;
  }
  cout << res << "\n";
  if (res) {
    for (int i : s) cout << i << " ";
    cout << "\n";
  }
  return 0;
}

```