#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
using namespace std;
const int N = 2e6 + 10;
const long long mod = 1e9 + 7;
const long long inf = 8e18;
const int LOG = 20;
long long pw(long long a, long long b, long long M = mod, long long ret = 1) {
  while (b) {
    ret = ret * (b & 1 ? a : 1) % M, a = a * a % M, b >>= 1;
  }
  return ret;
}
int n, m, ptr, sz, last[N], dp[N], in[N];
vector<pair<int, int> > vec[N];
vector<int> cm;
map<int, int> mp;
pair<int, int> seg[N << 2], lz[N << 2];
void upd(int l, int r, pair<int, int> x, int v = 1, int tl = 1, int tr = ptr) {
  if (l > tr || r < tl || l > r) return;
  if (l <= tl && tr <= r) {
    seg[v] = max(seg[v], x);
    lz[v] = max(lz[v], x);
    return;
  }
  int mid = (tl + tr) >> 1;
  upd(l, r, x, v << 1, tl, mid), upd(l, r, x, v << 1 | 1, mid + 1, tr);
  seg[v] = max(seg[v << 1], seg[v << 1 | 1]);
}
pair<int, int> get(int l, int r, int v = 1, int tl = 1, int tr = ptr) {
  if (l > tr || r < tl || l > r) return make_pair(0, 0);
  if (l <= tl && tr <= r) {
    return seg[v];
  }
  int mid = (tl + tr) >> 1;
  return max(
      {lz[v], get(l, r, v << 1, tl, mid), get(l, r, v << 1 | 1, mid + 1, tr)});
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int id, l, r;
    scanf("%d%d%d", &id, &l, &r);
    cm.push_back(l);
    cm.push_back(r);
    vec[id].push_back(make_pair(l, r));
  }
  cm.push_back(-1e9), cm.push_back(1e9);
  sort(cm.begin(), cm.end());
  cm.resize(unique(cm.begin(), cm.end()) - cm.begin());
  sz = (int)cm.size();
  for (int i = 0; i < sz; i++) {
    mp[cm[i]] = ++ptr;
    if (i + 1 != sz && cm[i] + 1 < cm[i + 1]) {
      ptr++;
    }
  }
  int Mx = 0;
  for (int i = 1; i <= n; i++) {
    dp[i] = 1;
    for (auto cu : vec[i]) {
      pair<int, int> calc = get(mp[cu.first], mp[cu.second]);
      if (calc.first + 1 > dp[i]) {
        dp[i] = calc.first + 1;
        last[i] = calc.second;
      }
    }
    for (auto cu : vec[i]) {
      upd(mp[cu.first], mp[cu.second], make_pair(dp[i], i));
    }
    if (dp[i] > dp[Mx]) Mx = i;
  }
  while (Mx > 0) {
    in[Mx] = 1;
    Mx = last[Mx];
  }
  vector<int> Ans;
  for (int i = 1; i <= n; i++) {
    if (in[i] == 0) Ans.push_back(i);
  }
  printf("%d\n", (int)Ans.size());
  for (auto x : Ans) {
    printf("%d ", x);
  }
  return 0;
}

### Prompt

Your challenge is to write 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>
#pragma GCC optimize("Ofast")
using namespace std;
const int N = 2e6 + 10;
const long long mod = 1e9 + 7;
const long long inf = 8e18;
const int LOG = 20;
long long pw(long long a, long long b, long long M = mod, long long ret = 1) {
  while (b) {
    ret = ret * (b & 1 ? a : 1) % M, a = a * a % M, b >>= 1;
  }
  return ret;
}
int n, m, ptr, sz, last[N], dp[N], in[N];
vector<pair<int, int> > vec[N];
vector<int> cm;
map<int, int> mp;
pair<int, int> seg[N << 2], lz[N << 2];
void upd(int l, int r, pair<int, int> x, int v = 1, int tl = 1, int tr = ptr) {
  if (l > tr || r < tl || l > r) return;
  if (l <= tl && tr <= r) {
    seg[v] = max(seg[v], x);
    lz[v] = max(lz[v], x);
    return;
  }
  int mid = (tl + tr) >> 1;
  upd(l, r, x, v << 1, tl, mid), upd(l, r, x, v << 1 | 1, mid + 1, tr);
  seg[v] = max(seg[v << 1], seg[v << 1 | 1]);
}
pair<int, int> get(int l, int r, int v = 1, int tl = 1, int tr = ptr) {
  if (l > tr || r < tl || l > r) return make_pair(0, 0);
  if (l <= tl && tr <= r) {
    return seg[v];
  }
  int mid = (tl + tr) >> 1;
  return max(
      {lz[v], get(l, r, v << 1, tl, mid), get(l, r, v << 1 | 1, mid + 1, tr)});
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int id, l, r;
    scanf("%d%d%d", &id, &l, &r);
    cm.push_back(l);
    cm.push_back(r);
    vec[id].push_back(make_pair(l, r));
  }
  cm.push_back(-1e9), cm.push_back(1e9);
  sort(cm.begin(), cm.end());
  cm.resize(unique(cm.begin(), cm.end()) - cm.begin());
  sz = (int)cm.size();
  for (int i = 0; i < sz; i++) {
    mp[cm[i]] = ++ptr;
    if (i + 1 != sz && cm[i] + 1 < cm[i + 1]) {
      ptr++;
    }
  }
  int Mx = 0;
  for (int i = 1; i <= n; i++) {
    dp[i] = 1;
    for (auto cu : vec[i]) {
      pair<int, int> calc = get(mp[cu.first], mp[cu.second]);
      if (calc.first + 1 > dp[i]) {
        dp[i] = calc.first + 1;
        last[i] = calc.second;
      }
    }
    for (auto cu : vec[i]) {
      upd(mp[cu.first], mp[cu.second], make_pair(dp[i], i));
    }
    if (dp[i] > dp[Mx]) Mx = i;
  }
  while (Mx > 0) {
    in[Mx] = 1;
    Mx = last[Mx];
  }
  vector<int> Ans;
  for (int i = 1; i <= n; i++) {
    if (in[i] == 0) Ans.push_back(i);
  }
  printf("%d\n", (int)Ans.size());
  for (auto x : Ans) {
    printf("%d ", x);
  }
  return 0;
}

```