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 t, n, i, j, k, x, y, z, l, r, mid, ans, mod = 1e9 + 7, g, m;
long long a[400005], b[400005], c[400005];
char ch;
string s, st;
vector<long long> v[300005], u;
int main() {
  ios::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  cin >> m;
  while (m--) {
    cin >> x;
    if (x == 1) {
      cin >> y >> k;
      for (i = 60; i >= 0; i--)
        if ((1LL << i) & y) break;
      a[i] = ((a[i] - k) % (1LL << i) + (1LL << i)) % (1LL << i);
    }
    if (x == 2) {
      cin >> y >> k;
      for (i = 60; i >= 0; i--)
        if ((1LL << i) & y) break;
      for (i; i <= 60; i++) {
        a[i] = ((a[i] - k) % (1LL << i) + (1LL << i)) % (1LL << i);
        k <<= 1;
      }
    }
    if (x == 3) {
      cin >> y;
      cout << y << " ";
      for (i = 60; i >= 0; i--)
        if ((1LL << i) & y) break;
      for (i = i - 1; i >= 0; i--) {
        z = y - a[i + 1];
        y = z / 2 + a[i];
        if (y >= (1LL << (i + 1))) y -= (1LL << i);
        if (y < (1LL << i)) y += (1LL << i);
        cout << y << " ";
      }
      cout << "\n";
    }
  }
}