#include <bits/stdc++.h>
using namespace std;
const int N = 62;
long long a[N], b[N];
long long mod[N];
int main() {
  mod[0] = 1;
  for (int i = 1; i < N; i++) {
    mod[i] = mod[i - 1] * 2;
  }
  int n, type;
  cin >> n;
  while (n--) {
    scanf("%d", &type);
    if (type == 1) {
      long long x, k;
      scanf("%lld%lld", &x, &k);
      int layer = log2(x);
      a[layer] = (a[layer] + k + mod[N - 1]) % mod[layer];
    }
    if (type == 2) {
      long long x, k;
      scanf("%lld%lld", &x, &k);
      int layer = log2(x);
      k = (k + mod[N - 1]) % mod[layer];
      for (int i = layer; i < 62; i++) {
        a[i] = (a[i] + k * mod[i - layer]) % mod[i];
      }
    }
    if (type == 3) {
      long long x;
      scanf("%lld", &x);
      int layer = log2(x);
      if (x == 1) {
        printf("%lld\n", x);
        continue;
      }
      printf("%lld ", x);
      b[layer] = x + a[layer] - mod[layer];
      for (int i = layer - 1; i >= 1; i--) {
        b[i] = (b[layer] / mod[layer - i] - a[i] + mod[i]) % mod[i];
      }
      for (int i = layer - 1; i >= 1; i--) {
        printf("%lld ", mod[i] + b[i]);
      }
      printf("%lld\n", 1 + b[0]);
    }
  }
  return 0;
}

### Prompt

Please formulate 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;
const int N = 62;
long long a[N], b[N];
long long mod[N];
int main() {
  mod[0] = 1;
  for (int i = 1; i < N; i++) {
    mod[i] = mod[i - 1] * 2;
  }
  int n, type;
  cin >> n;
  while (n--) {
    scanf("%d", &type);
    if (type == 1) {
      long long x, k;
      scanf("%lld%lld", &x, &k);
      int layer = log2(x);
      a[layer] = (a[layer] + k + mod[N - 1]) % mod[layer];
    }
    if (type == 2) {
      long long x, k;
      scanf("%lld%lld", &x, &k);
      int layer = log2(x);
      k = (k + mod[N - 1]) % mod[layer];
      for (int i = layer; i < 62; i++) {
        a[i] = (a[i] + k * mod[i - layer]) % mod[i];
      }
    }
    if (type == 3) {
      long long x;
      scanf("%lld", &x);
      int layer = log2(x);
      if (x == 1) {
        printf("%lld\n", x);
        continue;
      }
      printf("%lld ", x);
      b[layer] = x + a[layer] - mod[layer];
      for (int i = layer - 1; i >= 1; i--) {
        b[i] = (b[layer] / mod[layer - i] - a[i] + mod[i]) % mod[i];
      }
      for (int i = layer - 1; i >= 1; i--) {
        printf("%lld ", mod[i] + b[i]);
      }
      printf("%lld\n", 1 + b[0]);
    }
  }
  return 0;
}

```