#include <bits/stdc++.h>
using namespace std;
vector<pair<int, int> > vec[2102100];
pair<int, int> tr[2102100];
int lz[2102100];
int id[2102100];
int l[2102100];
int r[2102200];
int b[2102100];
int path[2102100];
int st[2102100];
void build(int p, int l, int r) {
  tr[p] = {0, -1};
  lz[p] = 0;
  if (l == r) {
    return;
  }
  int mid = l + r >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
}
void pushup(int p) { tr[p] = max(tr[p << 1], tr[p << 1 | 1]); }
void pushdown(int p) {
  if (lz[p]) {
    lz[p << 1] = lz[p << 1 | 1] = lz[p];
    lz[p] = 0;
    tr[p << 1] = tr[p << 1 | 1] = tr[p];
  }
}
pair<int, int> query(int p, int l, int r, int ll, int rr) {
  if (ll <= l && rr >= r) {
    return tr[p];
  }
  pushdown(p);
  int mid = l + r >> 1;
  pair<int, int> res = {0, -1};
  if (ll <= mid) res = max(query(p << 1, l, mid, ll, rr), res);
  if (rr > mid) res = max(query(p << 1 | 1, mid + 1, r, ll, rr), res);
  return res;
}
void update(int p, int l, int r, int ll, int rr, pair<int, int> res) {
  if (ll <= l && rr >= r) {
    tr[p] = res;
    lz[p] = 1;
    return;
  }
  pushdown(p);
  int mid = l + r >> 1;
  if (ll <= mid) update(p << 1, l, mid, ll, rr, res);
  if (rr > mid) update(p << 1 | 1, mid + 1, r, ll, rr, res);
  pushup(p);
}
int main() {
  int n, t;
  cin >> n >> t;
  for (int i = 1; i <= t; i++) {
    cin >> id[i] >> l[i] >> r[i];
    b[i * 2 - 1] = l[i];
    b[i * 2] = r[i];
  }
  sort(b + 1, b + 1 + t * 2);
  int len = unique(b + 1, b + 1 + t * 2) - b - 1;
  build(1, 1, len);
  for (int i = 1; i <= t; i++) {
    l[i] = lower_bound(b + 1, b + 1 + len, l[i]) - b;
    r[i] = lower_bound(b + 1, b + 1 + len, r[i]) - b;
    vec[id[i]].push_back({l[i], r[i]});
  }
  for (int i = 1; i <= n; i++) {
    pair<int, int> mx = {0, -1};
    for (auto tt : vec[i]) {
      mx = max(mx, query(1, 1, len, tt.first, tt.second));
    }
    mx.first++;
    path[i] = mx.second;
    mx.second = i;
    for (auto tt : vec[i]) {
      update(1, 1, len, tt.first, tt.second, mx);
    }
  }
  pair<int, int> ans = query(1, 1, len, 1, len);
  int u = ans.second;
  int aa = n;
  while (u != -1) {
    st[u] = 1;
    aa--;
    u = path[u];
  }
  cout << aa << endl;
  for (int i = 1; i <= n; i++) {
    if (!st[i]) cout << i << " ";
  }
}

### 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;
vector<pair<int, int> > vec[2102100];
pair<int, int> tr[2102100];
int lz[2102100];
int id[2102100];
int l[2102100];
int r[2102200];
int b[2102100];
int path[2102100];
int st[2102100];
void build(int p, int l, int r) {
  tr[p] = {0, -1};
  lz[p] = 0;
  if (l == r) {
    return;
  }
  int mid = l + r >> 1;
  build(p << 1, l, mid);
  build(p << 1 | 1, mid + 1, r);
}
void pushup(int p) { tr[p] = max(tr[p << 1], tr[p << 1 | 1]); }
void pushdown(int p) {
  if (lz[p]) {
    lz[p << 1] = lz[p << 1 | 1] = lz[p];
    lz[p] = 0;
    tr[p << 1] = tr[p << 1 | 1] = tr[p];
  }
}
pair<int, int> query(int p, int l, int r, int ll, int rr) {
  if (ll <= l && rr >= r) {
    return tr[p];
  }
  pushdown(p);
  int mid = l + r >> 1;
  pair<int, int> res = {0, -1};
  if (ll <= mid) res = max(query(p << 1, l, mid, ll, rr), res);
  if (rr > mid) res = max(query(p << 1 | 1, mid + 1, r, ll, rr), res);
  return res;
}
void update(int p, int l, int r, int ll, int rr, pair<int, int> res) {
  if (ll <= l && rr >= r) {
    tr[p] = res;
    lz[p] = 1;
    return;
  }
  pushdown(p);
  int mid = l + r >> 1;
  if (ll <= mid) update(p << 1, l, mid, ll, rr, res);
  if (rr > mid) update(p << 1 | 1, mid + 1, r, ll, rr, res);
  pushup(p);
}
int main() {
  int n, t;
  cin >> n >> t;
  for (int i = 1; i <= t; i++) {
    cin >> id[i] >> l[i] >> r[i];
    b[i * 2 - 1] = l[i];
    b[i * 2] = r[i];
  }
  sort(b + 1, b + 1 + t * 2);
  int len = unique(b + 1, b + 1 + t * 2) - b - 1;
  build(1, 1, len);
  for (int i = 1; i <= t; i++) {
    l[i] = lower_bound(b + 1, b + 1 + len, l[i]) - b;
    r[i] = lower_bound(b + 1, b + 1 + len, r[i]) - b;
    vec[id[i]].push_back({l[i], r[i]});
  }
  for (int i = 1; i <= n; i++) {
    pair<int, int> mx = {0, -1};
    for (auto tt : vec[i]) {
      mx = max(mx, query(1, 1, len, tt.first, tt.second));
    }
    mx.first++;
    path[i] = mx.second;
    mx.second = i;
    for (auto tt : vec[i]) {
      update(1, 1, len, tt.first, tt.second, mx);
    }
  }
  pair<int, int> ans = query(1, 1, len, 1, len);
  int u = ans.second;
  int aa = n;
  while (u != -1) {
    st[u] = 1;
    aa--;
    u = path[u];
  }
  cout << aa << endl;
  for (int i = 1; i <= n; i++) {
    if (!st[i]) cout << i << " ";
  }
}

```