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 nodeS[70], valueS[70];
long long po[70];
int getL(long long x) {
  int lvl = 1;
  while (x != 1) {
    x /= 2;
    lvl++;
  }
  return lvl;
}
int main() {
  po[1] = 1;
  for (int i = 2; i < 65; i++) {
    po[i] = po[i - 1] * 2;
  }
  int q;
  scanf("%d", &q);
  for (int i = 0; i < q; i++) {
    int t;
    long long x, k;
    scanf("%d %lld", &t, &x);
    if (t == 1) {
      scanf("%lld", &k);
      int level = getL(x);
      valueS[level] = (valueS[level] + (k % po[level]) + po[level]) % po[level];
    } else if (t == 2) {
      scanf("%lld", &k);
      int level = getL(x);
      nodeS[level] = (nodeS[level] + (k % po[level]) + po[level]) % po[level];
    } else {
      vector<long long> ans;
      long long cum = 0;
      long long mult = 2;
      int level = 1;
      stack<long long> st;
      while (mult <= x) {
        cum = (cum * 2) % po[level];
        cum += nodeS[level];
        st.push(cum + valueS[level]);
        level++;
        mult *= 2;
      }
      st.push((cum * 2 + valueS[level] + nodeS[level]) % po[level]);
      long long node =
          (x - po[level] + (st.top() % po[level]) + po[level]) % po[level] +
          po[level];
      if (node != 1) {
        ans.push_back(x);
        st.pop();
        node /= 2;
        level--;
      }
      while (node != 1) {
        long long back = st.top();
        st.pop();
        long long valueA =
            (node - po[level] - (back % po[level]) + po[level]) % po[level] +
            po[level];
        ans.push_back(valueA);
        node /= 2;
        level--;
      }
      ans.push_back(node);
      for (int i = 0; i < (int)ans.size(); i++) {
        printf("%lld ", ans[i]);
      }
      printf("\n");
    }
  }
  return 0;
}