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 N = 65;
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  x *= f;
}
long long ms[N];
long long dep[N];
void pre() {
  ms[0] = 1;
  for (int i = 1; i <= 60; i++) ms[i] = ms[i - 1] * 2;
}
int main() {
  int n;
  read(n);
  pre();
  while (n--) {
    int op;
    read(op);
    long long x, k;
    read(x);
    if (op < 3) read(k);
    int cy = log2(x);
    if (op == 1) {
      k %= ms[cy];
      dep[cy] += k;
      dep[cy] %= ms[cy];
    }
    if (op == 2) {
      k %= ms[cy];
      for (int i = cy; i <= 60; i++) (dep[i] += k * ms[i - cy]) %= ms[i];
    }
    if (op == 3) {
      long long now = x + dep[cy];
      now %= ms[cy];
      if (x == 1) {
        printf("1\n");
        continue;
      }
      printf("%I64d ", x);
      now /= 2;
      cy--;
      for (; cy >= 0; cy--) {
        long long tag = (ms[cy] - dep[cy] + now) % ms[cy];
        printf("%I64d ", tag + ms[cy]);
        now /= 2;
      }
      printf("\n");
    }
  }
}