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 IINF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fll;
const double DINF = numeric_limits<double>::infinity();
const int MOD = 1000000007;
const double EPS = 1e-9;
const double PI = acos(-1.0);
const int DX[] = {1, 0, -1, 0, 1, -1, 1, -1};
const int DY[] = {0, 1, 0, -1, 1, -1, -1, 1};
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long sqr(long long x) { return x * x; }
long long sqr(int x) { return (long long)x * x; }
double sqr(double x) { return x * x; }
long double sqr(long double x) { return x * x; }
mt19937 mmtw(960172);
long long rnd(long long x, long long y) {
  static uniform_int_distribution<long long> d;
  return d(mmtw) % (y - x + 1) + x;
}
const int N = 61;
long long st[N];
long long sh[N];
int getLev(long long x) {
  for (int i = 0; i < N; ++i) {
    if (x >= st[i] && (i == N - 1 || x < st[i + 1])) return i;
  }
  assert(0);
}
int main() {
  st[0] = 1;
  for (int i = 1; i < N; ++i) st[i] = st[i - 1] * 2;
  int qq;
  scanf("%d", &qq);
  while (qq--) {
    int tt;
    scanf("%d", &tt);
    if (tt == 3) {
      long long x;
      scanf("%lld", &x);
      int lev = getLev(x);
      long long pos = (x - st[lev] + sh[lev]) & ((1LL << lev) - 1);
      cout << x;
      while (lev != 0) {
        pos /= 2;
        --lev;
        cout << " " << (st[lev] + ((pos - sh[lev]) & ((1LL << lev) - 1)));
      }
      cout << "\n";
    } else {
      long long x, k;
      scanf("%lld%lld", &x, &k);
      int lev = getLev(x);
      k &= ((1LL << lev) - 1);
      for (int i = lev; i < (tt == 1 ? lev + 1 : N); ++i) {
        (sh[i] += k) &= ((1LL << i) - 1);
        k *= 2;
      }
    }
  }
  return 0;
}