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;
int q, type;
long long l[100], r[100], val[100], node[100], x, v;
int findDepth(long long x) {
  if (x == 1) return 0;
  for (int i = 1;; ++i) {
    if (r[i] >= x) return i;
  }
}
void shift(long long &x, long long k, int d) {
  long long n = r[d] - l[d] + 1;
  x += k;
  x -= l[d];
  x = (x % n + n) % n;
  x += l[d];
}
void shiftVal(long long x, long long k) {
  int d = findDepth(x);
  val[d] += k;
  val[d] %= (r[d] - l[d] + 1);
}
void shiftNode(long long x, long long k) {
  int d = findDepth(x);
  node[d] += k;
  node[d] %= (r[d] - l[d] + 1);
}
void print(long long x, int d) {
  if (!x) return;
  printf("%lld ", x);
  shift(x, val[d], d);
  shift(x, node[d], d);
  x /= 2;
  --d;
  shift(x, -val[d], d);
  print(x, d);
}
int main() {
  l[0] = r[0] = 1;
  for (int i = 1;; ++i) {
    l[i] = l[i - 1] * 2;
    r[i] = r[i - 1] * 2 + 1;
    if (r[i] > 1e18) {
      break;
    }
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d%lld", &type, &x);
    if (type == 3) {
      print(x, findDepth(x));
      puts("");
    } else {
      scanf("%lld", &v);
      if (type == 1) {
        shiftVal(x, v);
      } else {
        shiftNode(x, v);
      }
    }
  }
}