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 level[100];
long long levelLength[1000];
long long getLevel(long long x) {
  int l = 0;
  while (x) {
    x = x >> 1L;
    l++;
  }
  return l;
}
long long getLevelLength2(long long l) {
  long long ans = 1L;
  l--;
  while (l--) {
    ans *= 2L;
  }
  return ans;
}
long long getLevelLength(long long l) { return levelLength[l]; }
void shift(long long x, long long k, long long l) {
  long long length = getLevelLength(l);
  level[l] = (level[l] + k) % length;
}
void solve() {
  for (int i = 1; i <= 62; i++) {
    levelLength[i] = getLevelLength2(i);
  }
  int t;
  cin >> t;
  while (t--) {
    int type;
    cin >> type;
    if (type == 1) {
      long long x, k;
      cin >> x >> k;
      long long l = getLevel(x);
      long long length = getLevelLength(l);
      k %= length;
      k += length;
      k %= length;
      shift(x, k, l);
    } else if (type == 2) {
      long long x, k;
      cin >> x >> k;
      long long l = getLevel(x);
      long long length = getLevelLength(l);
      k %= length;
      k += length;
      k %= length;
      for (long long z = l; z <= 60; z++) {
        shift(x, k, l);
        x = x << 1L;
        k = k << 1L;
        l++;
      }
    } else {
      long long v;
      cin >> v;
      long long l = getLevel(v);
      l++;
      while (v != 1) {
        cout << v << ' ';
        l--;
        long long r2 = level[l];
        long long r1 = level[l - 1];
        long long diff = (r2 - 2 * r1);
        long long length = getLevelLength(l);
        diff %= length;
        diff += length;
        diff %= length;
        long long num = v + diff;
        long long lowerlimit = length;
        long long upperlimit = lowerlimit + lowerlimit - 1;
        if (num > upperlimit) {
          num = lowerlimit + (num - upperlimit) - 1;
        }
        v = num / 2;
      }
      cout << v << '\n';
    }
  }
}
int main() {
  cin.tie(0);
  cout.tie(0);
  cin.sync_with_stdio(0);
  cout.sync_with_stdio(0);
  int TC = 1;
  for (int ZZ = 1; ZZ <= TC; ZZ++) {
    clock_t start = clock();
    solve();
    clock_t end = clock();
    cerr << "Time: " << (double)(end - start) / CLOCKS_PER_SEC << " seconds"
         << '\n';
  }
  return 0;
}