#include <bits/stdc++.h>
using namespace std;
int intcmp(const void *v1, const void *v2) { return *(int *)v1 - *(int *)v2; }
long long minn(long long v1, long long v2) { return v1 < v2 ? v1 : v2; }
long long maxx(long long v1, long long v2) { return v1 > v2 ? v1 : v2; }
int Q;
long long er[100];
void init2() {
  er[0] = 1;
  for (int i = 1; i < 61; i++) er[i] = 2 * er[i - 1];
}
long long levelmovMUL2[100];
long long v2level(long long v) {
  for (int i = 0; i < 61; i++)
    if (v >= er[i] && v < er[i + 1]) return i;
  return 0;
}
void printans(long long v) {
  int level = v2level(v);
  printf("%lld ", v);
  if (v != 1) {
    printans(
        ((v + er[level] - levelmovMUL2[level - 1]) % (er[level]) + er[level]) /
        2);
  }
}
int main() {
  init2();
  cin >> Q;
  long long X, K;
  for (int i = 0; i < Q; i++) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      scanf("%lld %lld", &X, &K);
      int level = v2level(X);
      levelmovMUL2[level - 1] -= K;
      levelmovMUL2[level - 1] %= 2 * er[level];
      levelmovMUL2[level] += 2 * K;
      levelmovMUL2[level] %= 4 * er[level];
    } else if (t == 2) {
      scanf("%lld %lld", &X, &K);
      int level = v2level(X);
      levelmovMUL2[level - 1] -= K;
      levelmovMUL2[level - 1] %= 2 * er[level];
    } else if (t == 3) {
      scanf("%lld", &X);
      printans(X);
      cout << endl;
    }
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following 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>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int intcmp(const void *v1, const void *v2) { return *(int *)v1 - *(int *)v2; }
long long minn(long long v1, long long v2) { return v1 < v2 ? v1 : v2; }
long long maxx(long long v1, long long v2) { return v1 > v2 ? v1 : v2; }
int Q;
long long er[100];
void init2() {
  er[0] = 1;
  for (int i = 1; i < 61; i++) er[i] = 2 * er[i - 1];
}
long long levelmovMUL2[100];
long long v2level(long long v) {
  for (int i = 0; i < 61; i++)
    if (v >= er[i] && v < er[i + 1]) return i;
  return 0;
}
void printans(long long v) {
  int level = v2level(v);
  printf("%lld ", v);
  if (v != 1) {
    printans(
        ((v + er[level] - levelmovMUL2[level - 1]) % (er[level]) + er[level]) /
        2);
  }
}
int main() {
  init2();
  cin >> Q;
  long long X, K;
  for (int i = 0; i < Q; i++) {
    int t;
    scanf("%d", &t);
    if (t == 1) {
      scanf("%lld %lld", &X, &K);
      int level = v2level(X);
      levelmovMUL2[level - 1] -= K;
      levelmovMUL2[level - 1] %= 2 * er[level];
      levelmovMUL2[level] += 2 * K;
      levelmovMUL2[level] %= 4 * er[level];
    } else if (t == 2) {
      scanf("%lld %lld", &X, &K);
      int level = v2level(X);
      levelmovMUL2[level - 1] -= K;
      levelmovMUL2[level - 1] %= 2 * er[level];
    } else if (t == 3) {
      scanf("%lld", &X);
      printans(X);
      cout << endl;
    }
  }
  return 0;
}

```