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;
int lo(long long int a) {
  int i;
  for (i = 0; a > 1; i++) a /= 2;
  return i;
}
long long int po(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0)
    if (b % 2 == 1)
      ans *= a, b -= 1;
    else
      a *= a, b /= 2;
  return ans;
}
long long int lv[100];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      long long int x, k;
      cin >> x >> k;
      long long int level = lo(x);
      long long int co = po(2, level);
      long long int tmp = k;
      if (k < 0)
        k = (abs(k)) % co, k = -k;
      else
        k %= co;
      lv[level] += k;
      if (lv[level] < 0)
        lv[level] += co;
      else
        lv[level] %= co;
      if (level < 63) {
        k = tmp;
        k *= 2;
        co *= 2;
        if (k < 0)
          k = (abs(k)) % co, k = -k;
        else
          k %= co;
        lv[level + 1] -= k;
        if (lv[level + 1] < 0)
          lv[level + 1] += co, lv[level + 1] %= co;
        else
          lv[level + 1] %= (co);
      }
    } else if (t == 2) {
      long long int x, k;
      cin >> x >> k;
      long long int level = lo(x);
      long long int co = po(2, level);
      if (k < 0) k = (abs(k)) % co, k = co - k;
      k %= co;
      lv[level] += k;
      lv[level] %= co;
    } else {
      long long int x;
      cin >> x;
      long long int level = lo(x);
      vector<long long int> ans;
      while (x >= 1) {
        ans.push_back(x);
        x += lv[level];
        long long int co = po(2, level);
        x = (x - co) % co + co;
        level -= 1;
        x /= 2;
      }
      for (int i = 0; i < ans.size(); i++) cout << ans[i] << " ";
      cout << endl;
    }
  }
  return 0;
}