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 long long INF = 1e18;
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
inline long long add(long long x, long long y, long long mod) {
  x += y;
  x %= mod;
  if (x < 0) x += mod;
  return x;
}
inline long long fmul(int x, long long y, long long mod) {
  while (x) {
    y <<= 1;
    if (y >= mod) y -= mod;
    x -= 1;
  }
  return y;
}
long long fadd(long long x, long long y, long long mod) {
  x -= mod;
  x += y;
  x %= mod;
  if (x < 0) x += mod;
  x += mod;
  return x;
}
int f(long long x) { return 63 - __builtin_clzll(x); }
long long sizes[61], shifts[61];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  sizes[0] = 1;
  for (int i = 1; i <= 60; ++i) sizes[i] = sizes[i - 1] << 1;
  memset(shifts, 0, sizeof(shifts));
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      int lvl = f(x);
      shifts[lvl] = add(shifts[lvl], k, sizes[lvl]);
    } else if (t == 2) {
      long long x, k;
      cin >> x >> k;
      int lvl = f(x);
      int sto = lvl;
      while (lvl < 61) {
        shifts[lvl] =
            add(shifts[lvl], fmul(lvl - sto, k, sizes[lvl]), sizes[lvl]);
        lvl += 1;
      }
    } else {
      long long x;
      cin >> x;
      int lvl = f(x);
      cout << x << " ";
      x = add(x, shifts[lvl], sizes[lvl]);
      lvl -= 1;
      x >>= 1;
      while (lvl >= 0) {
        cout << fadd(x, -shifts[lvl], sizes[lvl]) << " ";
        x >>= 1;
        lvl -= 1;
      }
      cout << "\n";
    }
  }
  return 0;
}