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 Q;
long long query1[62], query2[62];
int main() {
  scanf("%d", &Q);
  for (int i = 0; i < Q; i++) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      long long x, k;
      scanf("%lld %lld", &x, &k);
      int p;
      for (int j = 0; j <= 60; j++) {
        if ((1LL << j) <= x && x < (1LL << (j + 1))) {
          p = j;
          break;
        }
      }
      query1[p] += k;
      query1[p] = (query1[p] % (1LL << p) + (1LL << p)) % (1LL << p);
    }
    if (t == 2) {
      long long x, k;
      scanf("%lld %lld", &x, &k);
      int p;
      for (int j = 0; j <= 60; j++) {
        if ((1LL << j) <= x && x < (1LL << (j + 1))) {
          p = j;
          break;
        }
      }
      query2[p] += k;
      query2[p] = (query2[p] % (1LL << p) + (1LL << p)) % (1LL << p);
    }
    if (t == 3) {
      long long x;
      scanf("%lld", &x);
      int p;
      for (int j = 0; j <= 60; j++) {
        if ((1LL << j) <= x && x < (1LL << (j + 1))) {
          p = j;
          break;
        }
      }
      while (x != 1) {
        printf("%lld ", x);
        long long y = x % (1LL << p);
        y += query1[p];
        y %= (1LL << p);
        y += query2[p];
        y %= (1LL << p);
        y += (1LL << p);
        y /= 2;
        y %= (1LL << (p - 1));
        y += (1LL << (p - 1)) - query1[p - 1];
        y %= (1LL << (p - 1));
        x = y + (1LL << (p - 1));
        p--;
      }
      printf("1\n");
    }
  }
}