#include <bits/stdc++.h>
using namespace std;
const int NMAX = 100005;
const long double eps = 1e-6;
long long shl[100];
int q;
long long log22(long long x) { return 64 - __builtin_clzll(x) - 1; }
int main() {
  ios::sync_with_stdio(false);
  scanf("%d", &q);
  long long t, x, k;
  while (q--) {
    scanf("%I64d%I64d", &t, &x);
    if (t == 1) {
      scanf("%I64d", &k);
      long long lvl = (long long)log22(x);
      shl[lvl] = (shl[lvl] + k) % (1LL << lvl);
      if (shl[lvl] < 0) shl[lvl] = (1LL << lvl) + shl[lvl];
    } else if (t == 2) {
      scanf("%I64d", &k);
      long long pow = 1;
      long long lvl = (long long)log22(x);
      for (int i = lvl; i <= 60; ++i, pow *= 2) {
        shl[i] =
            (shl[i] + ((k % (1LL << i)) * (pow % (1LL << i))) % (1LL << i)) %
            (1LL << i);
        if (shl[i] < 0) shl[i] = (1LL << i) + shl[i];
      }
    } else {
      printf("%I64d ", x);
      long long lvl;
      do {
        if (x == 1) break;
        lvl = (long long)log22(x);
        long long pos = x - (1LL << lvl) + 1;
        pos = (pos + shl[lvl]) % (1LL << lvl);
        if (pos == 0) pos = 1LL << lvl;
        x = (1LL << lvl) + pos - 1;
        long long tata = x / 2;
        --lvl;
        pos = tata - (1LL << lvl) + 1;
        pos = (pos + ((1LL << lvl) - shl[lvl])) % (1LL << lvl);
        if (pos == 0) pos = 1LL << lvl;
        tata = (1LL << lvl) + pos - 1;
        x = tata;
        printf("%I64d ", x);
      } while (x != 1);
      printf("\n");
    }
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 NMAX = 100005;
const long double eps = 1e-6;
long long shl[100];
int q;
long long log22(long long x) { return 64 - __builtin_clzll(x) - 1; }
int main() {
  ios::sync_with_stdio(false);
  scanf("%d", &q);
  long long t, x, k;
  while (q--) {
    scanf("%I64d%I64d", &t, &x);
    if (t == 1) {
      scanf("%I64d", &k);
      long long lvl = (long long)log22(x);
      shl[lvl] = (shl[lvl] + k) % (1LL << lvl);
      if (shl[lvl] < 0) shl[lvl] = (1LL << lvl) + shl[lvl];
    } else if (t == 2) {
      scanf("%I64d", &k);
      long long pow = 1;
      long long lvl = (long long)log22(x);
      for (int i = lvl; i <= 60; ++i, pow *= 2) {
        shl[i] =
            (shl[i] + ((k % (1LL << i)) * (pow % (1LL << i))) % (1LL << i)) %
            (1LL << i);
        if (shl[i] < 0) shl[i] = (1LL << i) + shl[i];
      }
    } else {
      printf("%I64d ", x);
      long long lvl;
      do {
        if (x == 1) break;
        lvl = (long long)log22(x);
        long long pos = x - (1LL << lvl) + 1;
        pos = (pos + shl[lvl]) % (1LL << lvl);
        if (pos == 0) pos = 1LL << lvl;
        x = (1LL << lvl) + pos - 1;
        long long tata = x / 2;
        --lvl;
        pos = tata - (1LL << lvl) + 1;
        pos = (pos + ((1LL << lvl) - shl[lvl])) % (1LL << lvl);
        if (pos == 0) pos = 1LL << lvl;
        tata = (1LL << lvl) + pos - 1;
        x = tata;
        printf("%I64d ", x);
      } while (x != 1);
      printf("\n");
    }
  }
  return 0;
}

```