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;
map<long long, long long> q1, q2;
pair<int, long long> conv(long long x) {
  int lvl = 0;
  long long mod = 1;
  while (x) {
    lvl++;
    x >>= 1;
    mod <<= 1;
  }
  return make_pair(lvl, mod / 2);
}
long long calc(long long x, long long mod, int lvl) {
  long long tmp = 0;
  if (q1.count(lvl)) {
    tmp += q1[lvl] + mod;
  }
  if (q2.count(lvl)) {
    tmp += q2[lvl] + mod;
  }
  return (x + tmp) % mod;
}
void solution() {
  int q;
  cin >> q;
  while (q--) {
    int t;
    long long x, k;
    scanf("%d%I64d", &t, &x);
    pair<int, long long> tmp = conv(x);
    int lvl = tmp.first;
    long long mod = tmp.second;
    if (t == 1) {
      scanf("%I64d", &k);
      q1[lvl] = (q1[lvl] + k) % mod;
    } else if (t == 2) {
      scanf("%I64d", &k);
      q2[lvl] = (q2[lvl] + k) % mod;
    } else {
      while (x > 1) {
        printf("%I64d ", x);
        long long pos = mod + calc(x, mod, lvl);
        x = (pos >> 1);
        mod >>= 1;
        lvl--;
        if (q1.count(lvl)) {
          x = mod + (x + mod - q1[lvl]) % mod;
        }
      }
      printf("1\n");
    }
  }
}
int main() { solution(); }
