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>
namespace loquat {
namespace bitmanip {
template <typename T>
inline auto ctz(T x) noexcept -> T {
  return __builtin_ctzll(x);
}
template <typename T>
inline auto clz(T x) noexcept -> T {
  return __builtin_clzll(x);
}
template <typename T>
inline T flp2(T x) noexcept {
  if (x == 0u) {
    return 0u;
  }
  return T(1u) << (sizeof(T) * 8u - 1u - clz(x));
}
}  // namespace bitmanip
}  // namespace loquat
using namespace std;
using ll = long long;
int main() {
  const int n = 62;
  vector<ll> shift(n);
  int q;
  scanf("%d", &q);
  while (q--) {
    int type;
    scanf("%d", &type);
    if (type == 1) {
      ll x, k;
      scanf("%lld%lld", &x, &k);
      const int t = loquat::bitmanip::ctz(loquat::bitmanip::flp2(x));
      k &= ((1ll << t) - 1);
      shift[t] = (shift[t] + k) & ((1ll << t) - 1);
    } else if (type == 2) {
      ll x, k;
      scanf("%lld%lld", &x, &k);
      const int t = loquat::bitmanip::ctz(loquat::bitmanip::flp2(x));
      k &= ((1ll << t) - 1);
      for (int i = t; i < n; ++i) {
        const ll ki = (k << (i - t));
        shift[i] = (shift[i] + ki) & ((1ll << i) - 1);
      }
    } else if (type == 3) {
      ll x;
      scanf("%lld", &x);
      const int t = loquat::bitmanip::ctz(loquat::bitmanip::flp2(x));
      x ^= (1ll << t);
      x = (x + shift[t]) & ((1ll << t) - 1);
      for (int i = t; i >= 0; --i) {
        const ll p = (x - shift[i]) & ((1ll << i) - 1);
        printf("%lld ", (p + (1ll << i)));
        x >>= 1;
      }
      printf("\n");
    }
  }
  return 0;
}