#include <bits/stdc++.h>
using namespace std;
long long sh[70];
int q;
int level(long long nod) {
  int biti = 0;
  while (nod) nod >>= 1, biti++;
  return biti;
}
int main() {
  scanf("%d", &q);
  while (q--) {
    int tip;
    scanf("%d", &tip);
    if (tip == 1) {
      long long x, lvl, val;
      scanf("%I64d%I64d", &x, &val);
      lvl = level(x);
      val %= (1LL << (lvl - 1));
      if (val < 0) val += (1LL << (lvl - 1));
      sh[lvl] += val;
      if (sh[lvl] >= (1LL << (lvl - 1))) sh[lvl] -= (1LL << (lvl - 1));
    } else if (tip == 2) {
      long long x, lvl, val;
      scanf("%I64d%I64d", &x, &val);
      lvl = level(x);
      val %= (1LL << (lvl - 1));
      if (val < 0) val += (1LL << (lvl - 1));
      for (int i = lvl; i < 63; i++) {
        sh[i] += val;
        sh[i] %= (1LL << (i - 1));
        val <<= 1;
      }
    } else {
      long long x;
      scanf("%I64d", &x);
      int lvl = level(x);
      x -= (1LL << (lvl - 1));
      x = (x + sh[lvl]) % (1LL << (lvl - 1));
      for (int i = lvl; i >= 1; i--) {
        long long r = x + (1LL << (i - 1)) - sh[i];
        if (r < 0) r += (1LL << (i - 1));
        if (r >= (1LL << (i - 1))) r -= (1LL << (i - 1));
        printf("%I64d ", r + (1LL << (i - 1)));
        x >>= 1;
      }
      printf("\n");
    }
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
long long sh[70];
int q;
int level(long long nod) {
  int biti = 0;
  while (nod) nod >>= 1, biti++;
  return biti;
}
int main() {
  scanf("%d", &q);
  while (q--) {
    int tip;
    scanf("%d", &tip);
    if (tip == 1) {
      long long x, lvl, val;
      scanf("%I64d%I64d", &x, &val);
      lvl = level(x);
      val %= (1LL << (lvl - 1));
      if (val < 0) val += (1LL << (lvl - 1));
      sh[lvl] += val;
      if (sh[lvl] >= (1LL << (lvl - 1))) sh[lvl] -= (1LL << (lvl - 1));
    } else if (tip == 2) {
      long long x, lvl, val;
      scanf("%I64d%I64d", &x, &val);
      lvl = level(x);
      val %= (1LL << (lvl - 1));
      if (val < 0) val += (1LL << (lvl - 1));
      for (int i = lvl; i < 63; i++) {
        sh[i] += val;
        sh[i] %= (1LL << (i - 1));
        val <<= 1;
      }
    } else {
      long long x;
      scanf("%I64d", &x);
      int lvl = level(x);
      x -= (1LL << (lvl - 1));
      x = (x + sh[lvl]) % (1LL << (lvl - 1));
      for (int i = lvl; i >= 1; i--) {
        long long r = x + (1LL << (i - 1)) - sh[i];
        if (r < 0) r += (1LL << (i - 1));
        if (r >= (1LL << (i - 1))) r -= (1LL << (i - 1));
        printf("%I64d ", r + (1LL << (i - 1)));
        x >>= 1;
      }
      printf("\n");
    }
  }
  return 0;
}

```