#include <bits/stdc++.h>
using namespace std;
inline void splay(long long &v) {
  v = 0;
  char c = 0;
  int p = 1;
  while (c < '0' || c > '9') {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    v = (v << 3) + (v << 1) + c - '0';
    c = getchar();
  }
  v *= p;
}
int Q;
long long v[66], a[66];
int main() {
  cin >> Q;
  v[0] = 1;
  for (int i = 1; i <= 60; i++) {
    v[i] = v[i - 1] * 2;
  }
  for (; Q--;) {
    long long id, x, k;
    splay(id);
    if (id == 1) {
      splay(x), splay(k);
      int tot = -1;
      for (long long i = x; i; i >>= 1) tot++;
      k = (k % v[tot] + v[tot]) % v[tot];
      a[tot] += k;
      a[tot] = (a[tot] % v[tot] + v[tot]) % v[tot];
    } else if (id == 2) {
      splay(x), splay(k);
      int tot = -1;
      for (long long i = x; i; i >>= 1) tot++;
      k = (k % v[tot] + v[tot]) % v[tot];
      for (int i = tot; i <= 60; i++) {
        a[i] += k;
        k <<= 1;
        a[i] = (a[i] % v[i] + v[i]) % v[i];
      }
    } else if (id == 3) {
      splay(x);
      int tot = -1;
      for (long long i = x; i; i >>= 1) tot++;
      x += a[tot];
      if (x >= v[tot] * 2) x -= v[tot];
      for (; x; x >>= 1, tot--) {
        long long ans = x - a[tot];
        if (ans < v[tot]) ans += v[tot];
        printf("%I64d ", ans);
      }
      puts("");
    }
  }
}

### 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;
inline void splay(long long &v) {
  v = 0;
  char c = 0;
  int p = 1;
  while (c < '0' || c > '9') {
    if (c == '-') p = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    v = (v << 3) + (v << 1) + c - '0';
    c = getchar();
  }
  v *= p;
}
int Q;
long long v[66], a[66];
int main() {
  cin >> Q;
  v[0] = 1;
  for (int i = 1; i <= 60; i++) {
    v[i] = v[i - 1] * 2;
  }
  for (; Q--;) {
    long long id, x, k;
    splay(id);
    if (id == 1) {
      splay(x), splay(k);
      int tot = -1;
      for (long long i = x; i; i >>= 1) tot++;
      k = (k % v[tot] + v[tot]) % v[tot];
      a[tot] += k;
      a[tot] = (a[tot] % v[tot] + v[tot]) % v[tot];
    } else if (id == 2) {
      splay(x), splay(k);
      int tot = -1;
      for (long long i = x; i; i >>= 1) tot++;
      k = (k % v[tot] + v[tot]) % v[tot];
      for (int i = tot; i <= 60; i++) {
        a[i] += k;
        k <<= 1;
        a[i] = (a[i] % v[i] + v[i]) % v[i];
      }
    } else if (id == 3) {
      splay(x);
      int tot = -1;
      for (long long i = x; i; i >>= 1) tot++;
      x += a[tot];
      if (x >= v[tot] * 2) x -= v[tot];
      for (; x; x >>= 1, tot--) {
        long long ans = x - a[tot];
        if (ans < v[tot]) ans += v[tot];
        printf("%I64d ", ans);
      }
      puts("");
    }
  }
}

```