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 power(long long _a, long long _b) {
  long long _r = 1;
  while (_b) {
    if (_b % 2 == 1) _r = (_r * _a);
    _b /= 2;
    _a = (_a * _a);
  }
  return _r;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  long long q, i, j;
  cin >> q;
  long long lvl[65];
  for (i = 0; i < 65; i++) lvl[i] = 0;
  while (q--) {
    long long t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      long long l = -1;
      while (x) {
        l++;
        x /= 2;
      }
      if (k < 0) {
        k = -k;
        k = k % power(2, l);
        k = -k;
      } else
        k = k % power(2, l);
      if (k > 0) k -= power(2, l);
      lvl[l] += k;
    } else if (t == 2) {
      long long x, k;
      cin >> x >> k;
      long long l = -1;
      while (x) {
        l++;
        x /= 2;
      }
      if (k < 0) {
        k = -k;
        k = k % power(2, l);
        k = -k;
      } else
        k = k % power(2, l);
      if (k > 0) k -= power(2, l);
      k = -k;
      for (i = l; i <= 60; i++) {
        lvl[i] -= k;
        lvl[i] = -lvl[i];
        lvl[i] = lvl[i] % power(2, i);
        lvl[i] = -lvl[i];
        k = k * 2;
      }
    } else {
      long long x;
      cin >> x;
      long long l = -1, o = x;
      while (o) {
        l++;
        o /= 2;
      }
      cout << x << " ";
      long long h = -lvl[l];
      h -= power(2, l);
      long long temp1 = x - h;
      long long g1 = power(2, l);
      temp1 = temp1 % g1;
      x = g1 + temp1;
      for (i = l - 1; i >= 0; i--) {
        x /= 2;
        long long temp = x - lvl[i];
        long long g = power(2, i);
        temp = temp % g;
        cout << g + temp << " ";
      }
      cout << endl;
    }
  }
  return 0;
}