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 shift[65];
long long two[65];
long long getlevel(long long node) {
  for (long long i = 1; i <= 60; i++) {
    if (two[i] + 1 > node) return i;
  }
}
void printshift() {
  for (int i = 1; i <= 4; i++) {
    printf("%lld ", shift[i]);
  }
  puts("");
}
int main() {
  two[0] = 1;
  for (long long i = 1; i <= 60; i++) {
    two[i] = two[i - 1] * 2;
  }
  for (long long i = 0; i <= 60; i++) {
    two[i]--;
  }
  int q;
  scanf("%d", &q);
  for (int _z = 0; _z < q; _z++) {
    int type;
    long long x, k;
    scanf("%d", &type);
    if (type == 1) {
      scanf("%lld %lld", &x, &k);
      long long L = getlevel(x);
      k += (two[60] + 1);
      k %= (two[L] - two[L - 1]);
      shift[L] += k;
      shift[L] %= (two[L] - two[L - 1]);
    } else if (type == 2) {
      scanf("%lld %lld", &x, &k);
      long long L = getlevel(x);
      k += (two[60] + 1);
      for (int i = L; i <= 60; i++) {
        k %= (two[i] - two[i - 1]);
        shift[i] += k;
        shift[i] %= (two[i] - two[i - 1]);
        k += k;
      }
    } else {
      scanf("%lld", &x);
      long long L = getlevel(x);
      x -= two[L - 1];
      x += shift[L];
      x %= (two[L] - two[L - 1]);
      if (x == 0) x = two[L] - two[L - 1];
      x += two[L - 1];
      for (int i = L; i >= 2; i--) {
        long long p = x;
        p -= two[i - 1];
        p -= shift[i];
        p += (two[i] - two[i - 1]);
        p %= (two[i] - two[i - 1]);
        if (p == 0) p = two[i] - two[i - 1];
        p += two[i - 1];
        printf("%lld ", p);
        x /= 2;
      }
      printf("1\n");
    }
  }
  return 0;
}