#include <bits/stdc++.h>
using namespace std;
const int maxn = 312345;
struct Seg {
  int l, r;
  bool lzy;
  pair<int, int> v;
} T[maxn << 3];
vector<pair<int, int> > G[maxn];
int nxt[maxn];
bool ans[maxn];
void build(int o, int l, int r) {
  T[o].l = l, T[o].r = r;
  T[o].v = {0, -1};
  if (l < r) {
    int mi = l + r >> 1;
    build(o << 1, l, mi);
    build(o << 1 | 1, mi + 1, r);
  }
}
void push_down(int o) {
  if (T[o].lzy) {
    T[o << 1].v = T[o << 1 | 1].v = T[o].v;
    T[o << 1].lzy = T[o << 1 | 1].lzy = true;
    T[o].lzy = false;
  }
}
pair<int, int> query(int o, int l, int r) {
  if (l <= T[o].l && T[o].r <= r) return T[o].v;
  push_down(o);
  int mi = T[o].l + T[o].r >> 1;
  pair<int, int> ret = {0, -1};
  if (l <= mi) ret = max(ret, query(o << 1, l, r));
  if (r > mi) ret = max(ret, query(o << 1 | 1, l, r));
  return ret;
}
void update(int o, int l, int r, pair<int, int> v) {
  if (l <= T[o].l && T[o].r <= r) {
    T[o].lzy = true;
    T[o].v = v;
    return;
  }
  int mi = T[o].l + T[o].r >> 1;
  if (l <= mi) update(o << 1, l, r, v);
  if (r > mi) update(o << 1 | 1, l, r, v);
  T[o].v = max(T[o << 1].v, T[o << 1 | 1].v);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  vector<int> tmp;
  for (int(i) = (1); (i) <= (m); (i)++) {
    int id, l, r;
    cin >> id >> l >> r;
    G[id].emplace_back(l, r);
    tmp.emplace_back(l);
    tmp.emplace_back(r);
  }
  sort((tmp).begin(), (tmp).end());
  (tmp).erase(unique((tmp).begin(), (tmp).end()), (tmp).end());
  auto loc = [&](int x) {
    return lower_bound((tmp).begin(), (tmp).end(), x) - tmp.begin() + 1;
  };
  for (int(i) = (1); (i) <= (n); (i)++)
    for (auto &[x, y] : G[i]) {
      x = loc(x);
      y = loc(y);
    }
  build(1, 1, tmp.size());
  for (int(i) = (1); (i) <= (n); (i)++) {
    pair<int, int> cur = {0, -1};
    for (auto [l, r] : G[i]) cur = max(cur, query(1, l, r));
    nxt[i] = cur.second;
    cur.first++, cur.second = i;
    for (auto [l, r] : G[i]) update(1, l, r, cur);
  }
  pair<int, int> mx = query(1, 1, tmp.size());
  cout << n - mx.first << '\n';
  for (int i = mx.second; i != -1; i = nxt[i]) ans[i] = true;
  for (int(i) = (1); (i) <= (n); (i)++)
    if (!ans[i]) cout << i << ' ';
}

### Prompt

Your task is to create 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 int maxn = 312345;
struct Seg {
  int l, r;
  bool lzy;
  pair<int, int> v;
} T[maxn << 3];
vector<pair<int, int> > G[maxn];
int nxt[maxn];
bool ans[maxn];
void build(int o, int l, int r) {
  T[o].l = l, T[o].r = r;
  T[o].v = {0, -1};
  if (l < r) {
    int mi = l + r >> 1;
    build(o << 1, l, mi);
    build(o << 1 | 1, mi + 1, r);
  }
}
void push_down(int o) {
  if (T[o].lzy) {
    T[o << 1].v = T[o << 1 | 1].v = T[o].v;
    T[o << 1].lzy = T[o << 1 | 1].lzy = true;
    T[o].lzy = false;
  }
}
pair<int, int> query(int o, int l, int r) {
  if (l <= T[o].l && T[o].r <= r) return T[o].v;
  push_down(o);
  int mi = T[o].l + T[o].r >> 1;
  pair<int, int> ret = {0, -1};
  if (l <= mi) ret = max(ret, query(o << 1, l, r));
  if (r > mi) ret = max(ret, query(o << 1 | 1, l, r));
  return ret;
}
void update(int o, int l, int r, pair<int, int> v) {
  if (l <= T[o].l && T[o].r <= r) {
    T[o].lzy = true;
    T[o].v = v;
    return;
  }
  int mi = T[o].l + T[o].r >> 1;
  if (l <= mi) update(o << 1, l, r, v);
  if (r > mi) update(o << 1 | 1, l, r, v);
  T[o].v = max(T[o << 1].v, T[o << 1 | 1].v);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, m;
  cin >> n >> m;
  vector<int> tmp;
  for (int(i) = (1); (i) <= (m); (i)++) {
    int id, l, r;
    cin >> id >> l >> r;
    G[id].emplace_back(l, r);
    tmp.emplace_back(l);
    tmp.emplace_back(r);
  }
  sort((tmp).begin(), (tmp).end());
  (tmp).erase(unique((tmp).begin(), (tmp).end()), (tmp).end());
  auto loc = [&](int x) {
    return lower_bound((tmp).begin(), (tmp).end(), x) - tmp.begin() + 1;
  };
  for (int(i) = (1); (i) <= (n); (i)++)
    for (auto &[x, y] : G[i]) {
      x = loc(x);
      y = loc(y);
    }
  build(1, 1, tmp.size());
  for (int(i) = (1); (i) <= (n); (i)++) {
    pair<int, int> cur = {0, -1};
    for (auto [l, r] : G[i]) cur = max(cur, query(1, l, r));
    nxt[i] = cur.second;
    cur.first++, cur.second = i;
    for (auto [l, r] : G[i]) update(1, l, r, cur);
  }
  pair<int, int> mx = query(1, 1, tmp.size());
  cout << n - mx.first << '\n';
  for (int i = mx.second; i != -1; i = nxt[i]) ans[i] = true;
  for (int(i) = (1); (i) <= (n); (i)++)
    if (!ans[i]) cout << i << ' ';
}

```