#include <bits/stdc++.h>
using namespace std;
void FAST_IO(string filein = "", string fileout = "") {
  if (fopen(filein.c_str(), "r")) {
    freopen(filein.c_str(), "r", stdin);
    freopen(fileout.c_str(), "w", stdout);
  }
  cin.tie(0), cout.tie(0)->sync_with_stdio(0);
}
void Why_Are_You_So_Orz();
signed main() {
  FAST_IO("mike4235orz.inp", "mike4235orz.out");
  int testcases = 1;
  for (int test = 1; test <= testcases; test++) Why_Are_You_So_Orz();
  return 0;
}
const int N = 3e5 + 5;
int n, m, dp[N], tr[N], cnt;
vector<pair<int, int>> q[N];
pair<int, int> st[N << 3], lz[N << 3];
void apply(int id, pair<int, int> v) {
  lz[id] = max(lz[id], v);
  st[id] = max(st[id], v);
}
void push(int id) { apply(id << 1, lz[id]), apply(id << 1 | 1, lz[id]); }
void update(int l, int r, pair<int, int> v, int id = 1, int tl = 1,
            int tr = cnt) {
  if (l > tr || tl > r) return;
  if (l <= tl && tr <= r) return apply(id, v);
  int tm = tl + tr >> 1;
  push(id);
  update(l, r, v, id << 1, tl, tm), update(l, r, v, id << 1 | 1, tm + 1, tr);
  st[id] = max(st[id << 1], st[id << 1 | 1]);
}
pair<int, int> query(int l, int r, int id = 1, int tl = 1, int tr = cnt) {
  if (l > tr || tl > r) return {0, 0};
  if (l <= tl && tr <= r) return st[id];
  int tm = tl + tr >> 1;
  push(id);
  return max(query(l, r, id << 1, tl, tm),
             query(l, r, id << 1 | 1, tm + 1, tr));
}
int chk[N];
void Why_Are_You_So_Orz() {
  cin >> n >> m;
  map<int, int> mp;
  for (int i = 1, v, l, r; i <= m; i++) {
    cin >> v >> l >> r, mp[l], mp[r], q[v].emplace_back(l, r);
  }
  for (auto &vp : mp) vp.second = ++cnt;
  for (int i = 1; i <= n; i++)
    for (auto &v : q[i]) v.first = mp[v.first], v.second = mp[v.second];
  for (int i = 1; i <= n; i++) {
    pair<int, int> tmp = {0, 0};
    for (auto lr : q[i]) tmp = max(tmp, query(lr.first, lr.second));
    dp[i] = tmp.first + 1, tr[i] = tmp.second;
    for (auto lr : q[i]) update(lr.first, lr.second, {dp[i], i});
  }
  int pos = max_element(dp + 1, dp + n + 1) - dp;
  while (pos) {
    chk[pos] = 1, pos = tr[pos];
  }
  vector<int> res;
  for (int i = 1; i <= n; i++)
    if (!chk[i]) res.push_back(i);
  cout << res.size() << '\n';
  for (auto v : res) cout << v << ' ';
}

### Prompt

Generate 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;
void FAST_IO(string filein = "", string fileout = "") {
  if (fopen(filein.c_str(), "r")) {
    freopen(filein.c_str(), "r", stdin);
    freopen(fileout.c_str(), "w", stdout);
  }
  cin.tie(0), cout.tie(0)->sync_with_stdio(0);
}
void Why_Are_You_So_Orz();
signed main() {
  FAST_IO("mike4235orz.inp", "mike4235orz.out");
  int testcases = 1;
  for (int test = 1; test <= testcases; test++) Why_Are_You_So_Orz();
  return 0;
}
const int N = 3e5 + 5;
int n, m, dp[N], tr[N], cnt;
vector<pair<int, int>> q[N];
pair<int, int> st[N << 3], lz[N << 3];
void apply(int id, pair<int, int> v) {
  lz[id] = max(lz[id], v);
  st[id] = max(st[id], v);
}
void push(int id) { apply(id << 1, lz[id]), apply(id << 1 | 1, lz[id]); }
void update(int l, int r, pair<int, int> v, int id = 1, int tl = 1,
            int tr = cnt) {
  if (l > tr || tl > r) return;
  if (l <= tl && tr <= r) return apply(id, v);
  int tm = tl + tr >> 1;
  push(id);
  update(l, r, v, id << 1, tl, tm), update(l, r, v, id << 1 | 1, tm + 1, tr);
  st[id] = max(st[id << 1], st[id << 1 | 1]);
}
pair<int, int> query(int l, int r, int id = 1, int tl = 1, int tr = cnt) {
  if (l > tr || tl > r) return {0, 0};
  if (l <= tl && tr <= r) return st[id];
  int tm = tl + tr >> 1;
  push(id);
  return max(query(l, r, id << 1, tl, tm),
             query(l, r, id << 1 | 1, tm + 1, tr));
}
int chk[N];
void Why_Are_You_So_Orz() {
  cin >> n >> m;
  map<int, int> mp;
  for (int i = 1, v, l, r; i <= m; i++) {
    cin >> v >> l >> r, mp[l], mp[r], q[v].emplace_back(l, r);
  }
  for (auto &vp : mp) vp.second = ++cnt;
  for (int i = 1; i <= n; i++)
    for (auto &v : q[i]) v.first = mp[v.first], v.second = mp[v.second];
  for (int i = 1; i <= n; i++) {
    pair<int, int> tmp = {0, 0};
    for (auto lr : q[i]) tmp = max(tmp, query(lr.first, lr.second));
    dp[i] = tmp.first + 1, tr[i] = tmp.second;
    for (auto lr : q[i]) update(lr.first, lr.second, {dp[i], i});
  }
  int pos = max_element(dp + 1, dp + n + 1) - dp;
  while (pos) {
    chk[pos] = 1, pos = tr[pos];
  }
  vector<int> res;
  for (int i = 1; i <= n; i++)
    if (!chk[i]) res.push_back(i);
  cout << res.size() << '\n';
  for (auto v : res) cout << v << ' ';
}

```