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 gcd(long long n1, long long n2) {
  if (n2 == 0) return n1;
  if (n1 % n2 == 0) return n2;
  return gcd(n2, n1 % n2);
}
long long powmod(long long base, long long exponent) {
  if (exponent < 0) exponent += 1000000007LL - 1;
  long long ans = 1;
  while (exponent) {
    if (exponent & 1) ans = (ans * base) % 1000000007LL;
    base = (base * base) % 1000000007LL;
    exponent /= 2;
  }
  return ans;
}
long long l[1000005];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  int q;
  cin >> q;
  while (q--) {
    int t;
    cin >> t;
    if (t == 1) {
      long long x, k;
      cin >> x >> k;
      long long n = x, cnt = 0;
      while (n > 0) {
        n /= 2LL;
        cnt++;
      }
      cnt--;
      k %= (1LL << cnt);
      if (k < 0) k += (1LL << cnt);
      l[cnt] += k;
      l[cnt] %= (1LL << cnt);
    } else if (t == 2) {
      long long x, k;
      cin >> x >> k;
      long long n = x, cnt = 0;
      while (n > 0) {
        n /= 2;
        cnt++;
      }
      cnt--;
      k %= (1LL << cnt);
      if (k < 0) k += (1LL << cnt);
      for (long long i = cnt; i < 60; i++) {
        k %= (1LL << i);
        l[i] += k;
        l[i] %= (1LL << i);
        k *= 2LL;
      }
    } else {
      long long x;
      cin >> x;
      cout << x << " ";
      if (x == 1) {
        cout << '\n';
        continue;
      }
      long long n = x, cnt = 0;
      while (n > 0) {
        n /= 2;
        cnt++;
      }
      cnt--;
      long long temp = x;
      long long temp2 = 1LL << cnt;
      temp = l[cnt];
      temp %= temp2;
      if (temp < 0) temp += temp2;
      temp += x;
      x = temp;
      x /= 2;
      cnt--;
      while (x != 1) {
        temp = x;
        temp2 = 1LL << cnt;
        temp = -l[cnt];
        temp %= temp2;
        if (temp < 0) temp += temp2;
        temp += x;
        while (temp >= temp2 * 2LL) temp -= temp2;
        while (temp < temp2) temp += temp2;
        if (temp == 1) break;
        cout << temp << " ";
        cnt--;
        x /= 2;
      }
      cout << 1 << '\n';
    }
  }
}