#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:640000000")
using namespace std;
const double eps = 1e-9;
const double pi = acos(-1.0);
long long a[111];
long long b[111];
long long st[111];
long long shift(long long x, long long y, long long mod) {
  x += y;
  x %= mod;
  x += mod;
  x %= mod;
  return x;
}
int get(long long x) {
  int res = 0;
  while (x > 0) {
    res++;
    x >>= 1;
  }
  return res - 1;
}
int main() {
  st[0] = 1;
  for (int i = 1; i < 63; i++) {
    st[i] = st[i - 1] * 2LL;
  }
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      int lvl = get(x);
      a[lvl] += k;
    } else if (t == 2) {
      long long x, k;
      cin >> x >> k;
      int lvl = get(x);
      b[lvl] += k;
      for (int i = lvl; i < 63; i++) {
        a[i] += k;
        k *= 2;
      }
    } else {
      long long x;
      cin >> x;
      int lvl = get(x);
      for (int i = lvl; i >= 0; i--) {
        printf("%I64d ", x);
        x = shift(x - st[i], a[i], st[i]) + st[i];
        x /= 2;
        if (i > 0) {
          x = shift(x - st[i - 1], -a[i - 1], st[i - 1]) + st[i - 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>
#pragma comment(linker, "/STACK:640000000")
using namespace std;
const double eps = 1e-9;
const double pi = acos(-1.0);
long long a[111];
long long b[111];
long long st[111];
long long shift(long long x, long long y, long long mod) {
  x += y;
  x %= mod;
  x += mod;
  x %= mod;
  return x;
}
int get(long long x) {
  int res = 0;
  while (x > 0) {
    res++;
    x >>= 1;
  }
  return res - 1;
}
int main() {
  st[0] = 1;
  for (int i = 1; i < 63; i++) {
    st[i] = st[i - 1] * 2LL;
  }
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      int lvl = get(x);
      a[lvl] += k;
    } else if (t == 2) {
      long long x, k;
      cin >> x >> k;
      int lvl = get(x);
      b[lvl] += k;
      for (int i = lvl; i < 63; i++) {
        a[i] += k;
        k *= 2;
      }
    } else {
      long long x;
      cin >> x;
      int lvl = get(x);
      for (int i = lvl; i >= 0; i--) {
        printf("%I64d ", x);
        x = shift(x - st[i], a[i], st[i]) + st[i];
        x /= 2;
        if (i > 0) {
          x = shift(x - st[i - 1], -a[i - 1], st[i - 1]) + st[i - 1];
        }
      }
      printf("\n");
    }
  }
  return 0;
}

```