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 n, i, j, k, x, y, z, m, q, h, r, timer, mx, l;
long long b[100500], a[1000500];
string s;
vector<long long> f;
long long find_level(long long x) {
  long long ans = -1;
  while (x) {
    x /= 2;
    ans++;
  }
  return ans;
}
int main() {
  cin >> q;
  while (q--) {
    long long test;
    cin >> test;
    if (test == 1) {
      cin >> x >> y;
      y = -y;
      long long lvl = find_level(x);
      a[lvl] += y;
      a[lvl] %= (1LL << lvl);
      while (a[lvl] < 0) {
        a[lvl] += (1LL << lvl);
      }
    } else if (test == 2) {
      cin >> x >> y;
      y = -y;
      long long lvl = find_level(x);
      y %= (1LL << lvl);
      if (y < 0) {
        y += (1LL << lvl);
      }
      for (; lvl < 61; lvl++) {
        a[lvl] += y;
        a[lvl] %= (1LL << lvl);
        while (a[lvl] < 0) {
          a[lvl] += (1LL << lvl);
        }
        y *= 2;
      }
    } else {
      cin >> x;
      long long lvl = find_level(x);
      x -= (1LL << lvl) + a[lvl];
      x %= (1LL << lvl);
      if (x < 0) {
        x += (1LL << lvl);
      }
      while (lvl >= 0) {
        cout << (1LL << lvl) + (x + a[lvl]) % (1LL << lvl) << " ";
        x /= 2;
        lvl--;
      }
      cout << endl;
    }
  }
  return 0;
}