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 long long LIMIT = 2e18;
long long shift[200];
int find_level(long long x) {
  int ans = 0;
  for (int level = 0;; level++) {
    if (1LL << level <= x) {
      ans = level;
    } else {
      break;
    }
  }
  return ans;
}
int main() {
  int q, typ;
  long long x, k;
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &typ);
    if (typ == 1) {
      scanf("%lld%lld", &x, &k);
      int level = find_level(x);
      long long mod = 1LL << level;
      shift[level] += k % mod;
      shift[level] %= mod;
    } else if (typ == 2) {
      scanf("%lld%lld", &x, &k);
      int level = find_level(x);
      for (; 1LL << level <= LIMIT; level++) {
        long long mod = 1LL << level;
        shift[level] += k % mod;
        shift[level] %= mod;
        k *= 2;
      }
    } else {
      scanf("%lld", &x);
      printf("%lld ", x);
      int level = find_level(x);
      long long mod = 1LL << level;
      long long actual = mod + (x + shift[level]) % mod;
      actual /= 2;
      level--;
      while (actual) {
        long long mod = 1LL << level;
        printf("%lld ", mod + (actual + mod - shift[level]) % mod);
        actual /= 2;
        level--;
      }
      printf("\n");
    }
  }
  return 0;
}