#include <bits/stdc++.h>
using namespace std;
long long shft[65];
int get_level(long long x) {
  int i = 0;
  while (x >= (1ll << i)) i++;
  return i;
}
void modify(long long &tmp, int lv) {
  tmp &= (1ll << lv) - 1;
  while (tmp > 1ll << lv) tmp -= 1ll << (lv - 1);
  while (tmp < 1ll << (lv)-1) tmp += 1ll << (lv - 1);
}
int main() {
  int i, j, k;
  int n, tp, lv;
  long long v, tmp, sh;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &tp);
    if (tp == 3) {
      scanf("%I64d", &v);
      lv = get_level(v);
      v = v - shft[lv];
      modify(v, lv);
      while (v > 0) {
        tmp = v + shft[lv];
        modify(tmp, lv);
        printf("%I64d ", tmp);
        v = v >> 1;
        lv--;
      }
      printf("\n");
    } else {
      scanf("%I64d%I64d", &v, &sh);
      lv = get_level(v);
      sh &= (1ll << (lv - 1)) - 1;
      shft[lv] -= sh;
      modify(shft[lv], lv);
      if (tp == 2) {
        lv++;
        while (lv <= 60) {
          sh <<= 1;
          shft[lv] -= sh;
          modify(shft[lv], lv);
          lv++;
        }
      }
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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;
long long shft[65];
int get_level(long long x) {
  int i = 0;
  while (x >= (1ll << i)) i++;
  return i;
}
void modify(long long &tmp, int lv) {
  tmp &= (1ll << lv) - 1;
  while (tmp > 1ll << lv) tmp -= 1ll << (lv - 1);
  while (tmp < 1ll << (lv)-1) tmp += 1ll << (lv - 1);
}
int main() {
  int i, j, k;
  int n, tp, lv;
  long long v, tmp, sh;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &tp);
    if (tp == 3) {
      scanf("%I64d", &v);
      lv = get_level(v);
      v = v - shft[lv];
      modify(v, lv);
      while (v > 0) {
        tmp = v + shft[lv];
        modify(tmp, lv);
        printf("%I64d ", tmp);
        v = v >> 1;
        lv--;
      }
      printf("\n");
    } else {
      scanf("%I64d%I64d", &v, &sh);
      lv = get_level(v);
      sh &= (1ll << (lv - 1)) - 1;
      shft[lv] -= sh;
      modify(shft[lv], lv);
      if (tp == 2) {
        lv++;
        while (lv <= 60) {
          sh <<= 1;
          shft[lv] -= sh;
          modify(shft[lv], lv);
          lv++;
        }
      }
    }
  }
  return 0;
}

```