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>
#include <bits/stdc++.h>
using namespace std;
const int inf = (1 << 30) - 1;
const long long linf = (1ll << 62) - 1;
int q;
map<int, long long> parent, value;
inline int getLevel(long long x) {
  int ans = 0;
  while (x > 1) {
    x >>= 1;
    ans++;
  }
  return ans;
}
int main() {
  cin >> q;
  for (int i = 0; i < q; i++) {
    int t;
    scanf("%d", &t);
    if (t == 1 || t == 2) {
      long long v, k;
      scanf("%lld%lld", &v, &k);
      int level = getLevel(v);
      if (t == 1) {
        value[level] = (value[level] + k + (1ll << level)) % (1ll << level);
      } else {
        parent[level] = (parent[level] + k + (1ll << level)) % (1ll << level);
      }
    } else {
      long long v;
      scanf("%lld", &v);
      int level = getLevel(v);
      long long start = (1ll << level);
      v = (v - start + value[level] + (2ll << level)) % (1ll << level) + start;
      for (int j = level; j >= 0; j--) {
        long long start = (1ll << j);
        printf("%lld ",
               (v - start - value[j] + (2ll << j)) % (1ll << j) + start);
        v = (v - start + parent[j] + (2ll << j)) % (1ll << j) + start;
        v >>= 1;
      }
      puts("");
    }
  }
  return 0;
}
