#include <bits/stdc++.h>
using namespace std;
const long double EPS = 1e-14;
const int maxn = 100500;
const int mod = 1e9 + 7;
long long lev[2][100];
int get(long long X) {
  if (X == 1) return 1;
  return 1 + get(X / 2);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> q;
  while (q--) {
    int t;
    long long x, k;
    cin >> t;
    if (t <= 2) {
      t--;
      cin >> x >> k;
      if (!t) k *= -1LL;
      int L = get(x);
      long long lo = (1ll << (L - 1)), hi = (1ll << L) - 1;
      long long len = hi - lo + 1;
      k %= len;
      lev[t][L] += k;
      lev[t][L] %= len;
    } else {
      cin >> x;
      vector<long long> res;
      res.push_back(x);
      int L = get(x);
      long long lo = (1ll << (L - 1)), hi = (1ll << L) - 1;
      long long len = hi - lo + 1;
      long long ad = x - lo;
      ad -= lev[0][L] - lev[1][L];
      ad = (ad + 3LL * len) % len;
      x = (ad + lo) >> 1;
      while (x) {
        int L = get(x);
        long long lo = (1ll << (L - 1)), hi = (1ll << L) - 1;
        long long len = hi - lo + 1;
        assert(x >= lo && x <= hi);
        long long deb = x;
        x -= lo;
        x += lev[0][L];
        x = (x + len) % len;
        res.push_back(x + lo);
        x = deb;
        x -= lo;
        x += lev[1][L];
        x = (x + len) % len;
        x = (x + lo) >> 1;
      }
      for (auto u : res) cout << u << ' ';
      cout << '\n';
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

You have a full binary tree having infinite levels.

Each node has an initial value. If a node has value x, then its left child has value 2·x and its right child has value 2·x + 1. 

The value of the root is 1. 

You need to answer Q queries. 

There are 3 types of queries: 

  1. Cyclically shift the values of all nodes on the same level as node with value X by K units. (The values/nodes of any other level are not affected).
  2. Cyclically shift the nodes on the same level as node with value X by K units. (The subtrees of these nodes will move along with them).
  3. Print the value of every node encountered on the simple path from the node with value X to the root.



Positive K implies right cyclic shift and negative K implies left cyclic shift. 

It is guaranteed that atleast one type 3 query is present.

Input

The first line contains a single integer Q (1 ≤ Q ≤ 105).

Then Q queries follow, one per line: 

  * Queries of type 1 and 2 have the following format: T X K (1 ≤ T ≤ 2; 1 ≤ X ≤ 1018; 0 ≤ |K| ≤ 1018), where T is type of the query.
  * Queries of type 3 have the following format: 3 X (1 ≤ X ≤ 1018).

Output

For each query of type 3, print the values of all nodes encountered in descending order.

Examples

Input

5
3 12
1 2 1
3 12
2 4 -1
3 8


Output

12 6 3 1 
12 6 2 1 
8 4 2 1 


Input

5
3 14
1 5 -3
3 14
1 3 1
3 14


Output

14 7 3 1 
14 6 3 1 
14 6 2 1 

Note

Following are the images of the first 4 levels of the tree in the first test case:

Original: 

<image>

After query 1 2 1: 

<image>

After query 2 4 -1: 

<image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const long double EPS = 1e-14;
const int maxn = 100500;
const int mod = 1e9 + 7;
long long lev[2][100];
int get(long long X) {
  if (X == 1) return 1;
  return 1 + get(X / 2);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q;
  cin >> q;
  while (q--) {
    int t;
    long long x, k;
    cin >> t;
    if (t <= 2) {
      t--;
      cin >> x >> k;
      if (!t) k *= -1LL;
      int L = get(x);
      long long lo = (1ll << (L - 1)), hi = (1ll << L) - 1;
      long long len = hi - lo + 1;
      k %= len;
      lev[t][L] += k;
      lev[t][L] %= len;
    } else {
      cin >> x;
      vector<long long> res;
      res.push_back(x);
      int L = get(x);
      long long lo = (1ll << (L - 1)), hi = (1ll << L) - 1;
      long long len = hi - lo + 1;
      long long ad = x - lo;
      ad -= lev[0][L] - lev[1][L];
      ad = (ad + 3LL * len) % len;
      x = (ad + lo) >> 1;
      while (x) {
        int L = get(x);
        long long lo = (1ll << (L - 1)), hi = (1ll << L) - 1;
        long long len = hi - lo + 1;
        assert(x >= lo && x <= hi);
        long long deb = x;
        x -= lo;
        x += lev[0][L];
        x = (x + len) % len;
        res.push_back(x + lo);
        x = deb;
        x -= lo;
        x += lev[1][L];
        x = (x + len) % len;
        x = (x + lo) >> 1;
      }
      for (auto u : res) cout << u << ' ';
      cout << '\n';
    }
  }
  return 0;
}

```