#include <bits/stdc++.h>
using namespace std;
long long p[71], level[71];
void powers() {
  p[0] = 1;
  for (int i = 1; i <= 61; i++) p[i] = p[i - 1] * 2LL;
}
void Update1(long long value, long long shift) {
  int lvl;
  for (int i = 61; i >= 0; i--) {
    if (p[i] <= value) {
      lvl = i;
      break;
    }
  }
  level[lvl] += shift + p[lvl];
  level[lvl] %= p[lvl];
}
void Update2(long long value, long long shift) {
  int lvl;
  for (int i = 61; i >= 0; i--) {
    if (p[i] <= value) {
      lvl = i;
      break;
    }
  }
  int j = 1;
  long long is = shift;
  for (int i = lvl; i <= 61; i++) {
    level[i] += shift + p[i];
    level[i] %= p[i];
    shift = (is * p[j]);
    j++;
  }
}
void PrintResult(long long value) {
  int lvl;
  for (int i = 61; i >= 0; i--) {
    if (p[i] <= value) {
      lvl = i;
      break;
    }
  }
  long long level_pos;
  printf("%lld", value);
  value = ((value - p[lvl] + level[lvl] + p[lvl]) % p[lvl]) + p[lvl];
  value /= 2LL;
  for (int i = lvl - 1; i >= 0; i--) {
    level_pos = value - p[i];
    level_pos = level_pos - level[i] + p[i];
    level_pos %= p[i];
    printf(" %lld", (p[i] + level_pos));
    value /= 2LL;
  }
  printf("\n");
}
int main() {
  powers();
  int q, type;
  long long shift, value;
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &type);
    if (type == 3) {
      scanf("%lld", &value);
      PrintResult(value);
    } else if (type == 2) {
      scanf("%lld%lld", &value, &shift);
      Update2(value, shift);
    } else {
      scanf("%lld%lld", &value, &shift);
      Update1(value, shift);
    }
  }
}

### 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;
long long p[71], level[71];
void powers() {
  p[0] = 1;
  for (int i = 1; i <= 61; i++) p[i] = p[i - 1] * 2LL;
}
void Update1(long long value, long long shift) {
  int lvl;
  for (int i = 61; i >= 0; i--) {
    if (p[i] <= value) {
      lvl = i;
      break;
    }
  }
  level[lvl] += shift + p[lvl];
  level[lvl] %= p[lvl];
}
void Update2(long long value, long long shift) {
  int lvl;
  for (int i = 61; i >= 0; i--) {
    if (p[i] <= value) {
      lvl = i;
      break;
    }
  }
  int j = 1;
  long long is = shift;
  for (int i = lvl; i <= 61; i++) {
    level[i] += shift + p[i];
    level[i] %= p[i];
    shift = (is * p[j]);
    j++;
  }
}
void PrintResult(long long value) {
  int lvl;
  for (int i = 61; i >= 0; i--) {
    if (p[i] <= value) {
      lvl = i;
      break;
    }
  }
  long long level_pos;
  printf("%lld", value);
  value = ((value - p[lvl] + level[lvl] + p[lvl]) % p[lvl]) + p[lvl];
  value /= 2LL;
  for (int i = lvl - 1; i >= 0; i--) {
    level_pos = value - p[i];
    level_pos = level_pos - level[i] + p[i];
    level_pos %= p[i];
    printf(" %lld", (p[i] + level_pos));
    value /= 2LL;
  }
  printf("\n");
}
int main() {
  powers();
  int q, type;
  long long shift, value;
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &type);
    if (type == 3) {
      scanf("%lld", &value);
      PrintResult(value);
    } else if (type == 2) {
      scanf("%lld%lld", &value, &shift);
      Update2(value, shift);
    } else {
      scanf("%lld%lld", &value, &shift);
      Update1(value, shift);
    }
  }
}

```