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 INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 1e5 + 5;
int sgn(double x) {
  if (fabs(x) < eps) return 0;
  if (x < 0)
    return -1;
  else
    return 1;
}
long long mv1[65], mv2[65];
int main() {
  int Q;
  scanf("%d", &Q);
  int op, dep;
  long long x, k, sz, mv;
  while (Q--) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%lld%lld", &x, &k);
      dep = 0;
      while (x > 0) {
        x /= 2;
        dep++;
      }
      sz = 1ll << (dep - 1);
      mv1[dep] = ((mv1[dep] + k) % sz + sz) % sz;
    } else if (op == 2) {
      scanf("%lld%lld", &x, &k);
      dep = 0;
      while (x > 0) {
        x /= 2;
        dep++;
      }
      sz = 1ll << (dep - 1);
      mv2[dep] = ((mv2[dep] + k) % sz + sz) % sz;
    } else {
      scanf("%lld", &x);
      long long t = x;
      dep = 0;
      while (t) {
        t /= 2;
        dep++;
      }
      while (x) {
        printf("%lld ", x);
        sz = 1ll << (dep - 1);
        mv = (mv1[dep] + mv2[dep]) % sz;
        x += mv;
        if (x >= 2 * sz) x -= sz;
        x /= 2;
        dep--;
        sz = 1ll << (dep - 1);
        x -= mv1[dep];
        if (x < sz) x += sz;
      }
      printf("\n");
    }
  }
}