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;
void init_ios() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
}
struct info {
  long long sft;
  long long mn, mx;
};
int main() {
  init_ios();
  int Q;
  cin >> Q;
  vector<info> t(61);
  for (int i = 1; i <= 60; ++i) {
    t[i].mn = 1LL << (i - 1);
    t[i].mx = (1LL << i) - 1;
  }
  auto getLevel = [&](long long x) {
    for (int l = 1; l <= 60; ++l) {
      if (x <= t[l].mx) return l;
    }
    return -1;
  };
  for (int q = 0; q < Q; ++q) {
    int T;
    cin >> T;
    if (T < 3) {
      long long X, K;
      cin >> X >> K;
      if (T == 1) {
        int l = getLevel(X);
        (t[l].sft += K) %= t[l].mn;
      } else {
        long long k = K;
        for (int l = getLevel(X); l <= 60; ++l) {
          (t[l].sft += k) %= t[l].mn;
          (k *= 2) %= 1LL << 61;
        }
      }
    } else {
      long long X;
      cin >> X;
      int l = getLevel(X);
      vector<long long> res{X};
      long long p = t[l].mn + (((X + t[l].sft) % t[l].mn + t[l].mn) % t[l].mn);
      p /= 2;
      while (0 < --l) {
        long long x =
            t[l].mn + (((p - t[l].sft) % t[l].mn + t[l].mn) % t[l].mn);
        res.push_back(x);
        p /= 2;
      }
      for (int i = 0; i < res.size(); ++i) {
        if (i) cout << " ";
        cout << res[i];
      }
      cout << endl;
    }
  }
  return 0;
}