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;
long long int pw[61];
long long int offset[61];
long long int modu(long long int x) {
  if (x < 0) return (-x);
  return x;
}
long long int signum(long long int x) {
  if (x < 0) return -1;
  if (x > 0) return 1;
  return 0;
}
int findLevel(long long int x) {
  int res = 0;
  while (x > 1) {
    x /= 2;
    res++;
  }
  return res;
}
long long int findPos(int h, long long int x) {
  long long int pos = pw[h] + ((x - pw[h] - offset[h]) % pw[h] + pw[h]) % pw[h];
  return pos;
}
void printAnc(long long int x) {
  long long int pos;
  int h = findLevel(x);
  pos = pw[h] + ((x - pw[h] + offset[h]) % pw[h] + pw[h]) % pw[h];
  while (h >= 0) {
    long long int y = findPos(h, pos);
    h--;
    pos /= 2;
    printf("%lld ", y);
  }
  printf("\n");
}
int main() {
  int q, t, h;
  long long int x, k;
  pw[0] = 1;
  for (int i = 1; i < 61; i++) pw[i] = pw[i - 1] * 2;
  cin >> q;
  memset(offset, 0, sizeof(offset));
  for (int i = 0; i < q; i++) {
    scanf("%d", &t);
    switch (t) {
      case 1:
        scanf("%lld %lld", &x, &k);
        h = findLevel(x);
        offset[h] += (signum(k) * ((modu(k) % pw[h])));
        offset[h] = signum(offset[h]) * ((modu(offset[h]) % pw[h]));
        break;
      case 2:
        scanf("%lld %lld", &x, &k);
        h = findLevel(x);
        for (int j = h; j < 61; j++) {
          k = signum(k) * (modu(k) % pw[j]);
          offset[j] += k;
          offset[h] = signum(offset[h]) * ((modu(offset[h]) % pw[h]));
          if (j != 60) k *= 2;
        }
        break;
      case 3:
        scanf("%lld", &x);
        printAnc(x);
        break;
    }
  }
  return 0;
}