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;
const long long mo = 1e9 + 7;
int q;
long long x, k;
long long shift[100];
long long st[100];
int main() {
  cin >> q;
  st[0] = 1;
  for (int i = 1; i <= 60; i++) st[i] = st[i - 1] * 2;
  while (q--) {
    int tip;
    scanf("%d", &tip);
    if (tip == 1) {
      scanf("%lld%lld", &x, &k);
      long long last = 0;
      for (long long j = 0; j <= 60; j++)
        if (x & (1ll << j)) last = j;
      shift[last] += k;
      shift[last] %= st[last];
      shift[last] += st[last];
      shift[last] %= st[last];
    } else if (tip == 2) {
      scanf("%lld%lld", &x, &k);
      long long last = 0;
      for (long long j = 0; j <= 60; j++)
        if (x & (1ll << j)) last = j;
      long long cur = 1;
      for (long long j = last; j <= 60; j++) {
        long long val = k;
        for (int l = 1; l < cur; l++) {
          val *= 2;
          val %= st[j];
        }
        shift[j] += val;
        shift[j] %= st[j];
        shift[j] += st[j];
        shift[j] %= st[j];
        cur++;
      }
    } else {
      scanf("%lld", &x);
      long long last = 0;
      for (long long j = 0; j <= 60; j++)
        if (x & (1ll << j)) last = j;
      for (int j = last; j >= 0; j--) {
        if (j != last) {
          long long p = x;
          p -= shift[j];
          if (p < st[j]) p += st[j];
          printf("%lld ", p);
          x /= 2;
        } else {
          printf("%lld ", x);
          long long p = x;
          p += shift[j];
          if (p >= st[j + 1]) p -= st[j];
          x = p / 2;
        }
      }
      printf("\n");
    }
  }
  return 0;
}