#include <bits/stdc++.h>
using namespace std;
const int maxn = 60 + 10;
const double EPS = 1e-6;
long long shi[maxn];
map<int, long long> cs;
inline long long GetMod(long long a, long long mod) {
  if (a >= mod)
    return a - mod;
  else if (a < 0)
    return a + mod;
  else
    return a;
}
int main() {
  int q;
  for (int i = 0; i <= 60; i++) {
    cs[i] = (long long)(pow(2, i) + EPS);
  }
  while (scanf("%d", &q) != EOF) {
    memset(shi, 0, sizeof shi);
    for (int i = 0; i < q; i++) {
      long long t, x, k;
      scanf("%lld%lld", &t, &x);
      int h = 64 - __builtin_clzll(x);
      if (t == 3) {
        int l = h;
        printf("%lld ", x);
        long long tmp = cs[l - 1];
        long long k = (tmp + ((x - tmp + shi[l] + tmp + tmp) % tmp)) / 2;
        for (l -= 1; l; l--, k /= 2) {
          tmp = cs[l - 1];
          printf("%lld ", tmp + (((k - tmp - shi[l]) + tmp + tmp) % tmp));
        }
        printf("\n");
      } else {
        scanf("%lld", &k);
        long long tmp = cs[h - 1];
        k %= tmp;
        if (k < 0) k += tmp;
        if (t == 1) {
          shi[h] = (shi[h] + k + tmp) % tmp;
        } else if (t == 2) {
          for (int l = h; l <= 60; l++) {
            long long n_w = cs[l - 1];
            shi[l] = GetMod(shi[l] + GetMod(k * cs[l - h], n_w), n_w);
          }
        }
      }
    }
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int maxn = 60 + 10;
const double EPS = 1e-6;
long long shi[maxn];
map<int, long long> cs;
inline long long GetMod(long long a, long long mod) {
  if (a >= mod)
    return a - mod;
  else if (a < 0)
    return a + mod;
  else
    return a;
}
int main() {
  int q;
  for (int i = 0; i <= 60; i++) {
    cs[i] = (long long)(pow(2, i) + EPS);
  }
  while (scanf("%d", &q) != EOF) {
    memset(shi, 0, sizeof shi);
    for (int i = 0; i < q; i++) {
      long long t, x, k;
      scanf("%lld%lld", &t, &x);
      int h = 64 - __builtin_clzll(x);
      if (t == 3) {
        int l = h;
        printf("%lld ", x);
        long long tmp = cs[l - 1];
        long long k = (tmp + ((x - tmp + shi[l] + tmp + tmp) % tmp)) / 2;
        for (l -= 1; l; l--, k /= 2) {
          tmp = cs[l - 1];
          printf("%lld ", tmp + (((k - tmp - shi[l]) + tmp + tmp) % tmp));
        }
        printf("\n");
      } else {
        scanf("%lld", &k);
        long long tmp = cs[h - 1];
        k %= tmp;
        if (k < 0) k += tmp;
        if (t == 1) {
          shi[h] = (shi[h] + k + tmp) % tmp;
        } else if (t == 2) {
          for (int l = h; l <= 60; l++) {
            long long n_w = cs[l - 1];
            shi[l] = GetMod(shi[l] + GetMod(k * cs[l - h], n_w), n_w);
          }
        }
      }
    }
  }
  return 0;
}

```