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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long f[61 + 1], g[61 + 1], x, k;
int Q, op, pos;
int main() {
  scanf("%d", &Q);
  f[0] = 1;
  for (int i = 1; i < 61; i++) f[i] = f[i - 1] * 2;
  while (Q--) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%I64d%I64d", &x, &k);
      for (int i = 0; i < 61; i++)
        if (f[i] >= x) {
          pos = i;
          break;
        } else
          x -= f[i];
      g[pos] = (g[pos] + k) % f[pos];
    }
    if (op == 2) {
      scanf("%I64d%I64d", &x, &k);
      for (int i = 0; i < 61; i++)
        if (f[i] >= x) {
          pos = i;
          break;
        } else
          x -= f[i];
      k %= f[pos];
      for (int i = pos; i < 61; i++) {
        g[i] = (g[i] + k) % f[i];
        k *= 2;
      }
    }
    if (op == 3) {
      scanf("%I64d", &x);
      printf("%I64d", x);
      long long tmp = x;
      for (int i = 0; i < 61; i++)
        if (f[i] >= x) {
          pos = i;
          break;
        } else
          x -= f[i];
      x = tmp;
      while (x != 1) {
        long long xx = x, fa;
        xx += g[pos];
        if (xx < f[pos]) xx += f[pos];
        if (xx >= f[pos + 1]) xx -= f[pos];
        fa = xx / 2;
        pos--;
        fa -= g[pos];
        if (fa < f[pos]) fa += f[pos];
        if (fa >= f[pos + 1]) fa -= f[pos];
        x = fa;
        printf(" %I64d", fa);
      }
      printf("\n");
    }
  }
}