#include <bits/stdc++.h>
using namespace std;
vector<long long> pw;
int bitcnt(long long x) {
  for (int i = 0; i <= 60; ++i) {
    if (x <= pw[i + 1] - 1) return i;
  }
}
int main() {
  int Q;
  scanf("%d", &Q);
  vector<long long> shifts(61);
  pw.resize(62);
  pw[0] = 1;
  for (int i = 1; i <= 61; ++i) pw[i] = pw[i - 1] * 2;
  for (int ip = 0; ip < Q; ++ip) {
    int t;
    long long x, k;
    scanf("%d %lld", &t, &x);
    if (t != 3) scanf("%lld", &k);
    k = (k + pw[61]) % pw[61];
    if (t == 1) {
      int y = bitcnt(x);
      shifts[y] = (shifts[y] + k) & (pw[y] - 1);
    }
    if (t == 2) {
      int y = bitcnt(x);
      for (int i = y; i < 61; ++i) {
        shifts[i] = (shifts[i] + k) & (pw[i] - 1);
        k = (k << 1) & (pw[61] - 1);
      }
    }
    if (t == 3) {
      int y = bitcnt(x);
      long long pos = (x + shifts[y]) & (pw[y] - 1);
      for (int i = y; i >= 0; --i) {
        long long val = ((pos + pw[i] - shifts[i]) & (pw[i] - 1)) + pw[i];
        printf("%lld ", val);
        pos >>= 1;
      }
      printf("\n");
    }
  }
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

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;
vector<long long> pw;
int bitcnt(long long x) {
  for (int i = 0; i <= 60; ++i) {
    if (x <= pw[i + 1] - 1) return i;
  }
}
int main() {
  int Q;
  scanf("%d", &Q);
  vector<long long> shifts(61);
  pw.resize(62);
  pw[0] = 1;
  for (int i = 1; i <= 61; ++i) pw[i] = pw[i - 1] * 2;
  for (int ip = 0; ip < Q; ++ip) {
    int t;
    long long x, k;
    scanf("%d %lld", &t, &x);
    if (t != 3) scanf("%lld", &k);
    k = (k + pw[61]) % pw[61];
    if (t == 1) {
      int y = bitcnt(x);
      shifts[y] = (shifts[y] + k) & (pw[y] - 1);
    }
    if (t == 2) {
      int y = bitcnt(x);
      for (int i = y; i < 61; ++i) {
        shifts[i] = (shifts[i] + k) & (pw[i] - 1);
        k = (k << 1) & (pw[61] - 1);
      }
    }
    if (t == 3) {
      int y = bitcnt(x);
      long long pos = (x + shifts[y]) & (pw[y] - 1);
      for (int i = y; i >= 0; --i) {
        long long val = ((pos + pw[i] - shifts[i]) & (pw[i] - 1)) + pw[i];
        printf("%lld ", val);
        pos >>= 1;
      }
      printf("\n");
    }
  }
  return 0;
}

```