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 int N = 1e3 + 7;
const int M = 5e5 + 5;
const long long mod = 998244353;
const long long inf = 0x3f3f3f3f;
const int limit = 63;
long long markv[limit];
long long markp[limit];
long long posnode[limit];
long long findrightpos(int high, long long v) {
  if (!high) return 0;
  long long now = v - posnode[high];
  long long cut = (posnode[high] + markv[high]) % posnode[high];
  now = (now + cut) % posnode[high];
  return now + posnode[high];
}
long long findrightval(int high, long long v) {
  if (!high) return 0;
  long long now = v - posnode[high];
  long long cut = ((posnode[high] - markv[high])) % posnode[high];
  now = (now + cut) % posnode[high];
  return now + posnode[high];
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  posnode[1] = 1;
  for (int i = 2; i < limit; i++) posnode[i] = posnode[i - 1] * 2;
  int q;
  cin >> q;
  while (q--) {
    int o;
    long long x, k;
    cin >> o;
    if (o == 1) {
      cin >> x >> k;
      int h = 0;
      long long tmp = x;
      for (int i = 0; i < limit; i++) {
        if (tmp <= posnode[i]) {
          h = i;
          break;
        } else
          tmp -= posnode[i];
      }
      x = h;
      if (x >= limit)
        continue;
      else {
        k %= posnode[x];
        k += posnode[x];
        k %= posnode[x];
        markv[x] = (markv[x] + k) % posnode[x];
      }
    } else if (o == 2) {
      cin >> x >> k;
      int h = 0;
      long long tmp = x;
      for (int i = 0; i < limit; i++) {
        if (tmp <= posnode[i]) {
          h = i;
          break;
        } else
          tmp -= posnode[i];
      }
      x = h;
      if (x >= limit)
        continue;
      else {
        k %= posnode[x];
        k += posnode[x];
        k %= posnode[x];
        for (int j = x; j < limit; j++) {
          long long t = k;
          for (int z = 0; z < (j - x); z++) t = t * 2 % posnode[j];
          markv[j] = (markv[j] + t) % posnode[j];
        }
      }
    } else {
      long long now;
      cin >> now;
      int h = 0;
      long long tmp = now;
      for (int i = 0; i < limit; i++) {
        if (tmp <= posnode[i]) {
          h = i;
          break;
        } else
          tmp -= posnode[i];
      }
      for (int f = h; f >= 1; f--) {
        cout << now << " ";
        now = findrightpos(f, now);
        now = findrightval(f - 1, now / 2);
      }
      cout << endl;
    }
  }
}