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 x, k, val, val1, power[65], res[65];
int main() {
  int q, i, typ, j;
  scanf("%d", &q);
  for (i = 0; i < 64; i++) {
    power[i] = i > 0 ? power[i - 1] * 2LL : 1;
  }
  for (i = 0; i < q; i++) {
    scanf("%d", &typ);
    if (typ == 1) {
      scanf("%lld%lld", &x, &k);
      for (j = 0; j < 65; j++) {
        if (x / power[j] == 0) {
          break;
        }
      }
      val = (j - 1);
      val++;
      res[val] += k;
      res[val] %= power[val - 1];
      if (res[val] < 0) {
        res[val] += power[val - 1];
        res[val] %= power[val - 1];
      }
    } else if (typ == 2) {
      scanf("%lld%lld", &x, &k);
      for (j = 0; j < 65; j++) {
        if (x / power[j] == 0) {
          break;
        }
      }
      val = (j - 1);
      val++;
      for (j = val; j < 64; j++) {
        res[j] += (k % power[j - 1]) * power[j - val];
        res[j] %= power[j - 1];
        if (res[j] < 0) {
          res[j] += power[j - 1];
          res[j] %= power[j - 1];
        }
      }
    } else {
      scanf("%lld", &x);
      for (j = 0; j < 65; j++) {
        if (x / power[j] == 0) {
          break;
        }
      }
      val = (j - 1);
      val++;
      printf("%lld ", x);
      if (x + res[val] < power[val]) {
        x = x + res[val];
      } else {
        val1 = res[val];
        val1 -= (power[val] - x);
        x = power[val - 1] + val1;
      }
      x /= 2;
      long long x1;
      while (x > 0) {
        val--;
        if (power[val - 1] + res[val] <= x) {
          x1 = x - res[val];
        } else {
          x1 = power[val] - (res[val] + power[val - 1] - x);
        }
        printf("%lld ", x1);
        x /= 2;
      }
      printf("\n");
    }
  }
}