#include <bits/stdc++.h>
const int inf = (1ll << 30) - 1;
const int maxn = (int)1e5 + 10;
using namespace std;
long long L[111], R[111], SHIFT[111], SHIFTVAL[111];
int sz = 1;
void upd1() {
  long long x, k;
  scanf("%lld%lld", &x, &k);
  int g = 1;
  while (R[g] < x) g++;
  SHIFTVAL[g] -= k;
  SHIFTVAL[g] %= (R[g] - L[g] + 1);
  SHIFTVAL[g] += (R[g] - L[g] + 1);
  SHIFTVAL[g] %= (R[g] - L[g] + 1);
}
void upd2() {
  long long x, k;
  scanf("%lld%lld", &x, &k);
  int g = 1;
  while (R[g] < x) g++;
  while (g <= sz) {
    SHIFTVAL[g] -= k;
    SHIFTVAL[g] %= (R[g] - L[g] + 1);
    SHIFTVAL[g] += (R[g] - L[g] + 1);
    SHIFTVAL[g] %= (R[g] - L[g] + 1);
    k = (k * 2) % (R[sz] - L[sz] + 1);
    g++;
  }
}
void get() {
  long long x;
  scanf("%lld", &x);
  int g = 1;
  while (R[g] < x) g++;
  long long pos = x - L[g];
  pos -= SHIFTVAL[g];
  if (pos < 0) pos += R[g] - L[g] + 1;
  while (g > 0) {
    long long d = (SHIFTVAL[g] + pos) % (R[g] - L[g] + 1);
    printf("%lld", d + L[g]);
    pos /= 2;
    if (g > 1) printf(" ");
    g--;
  }
  printf("\n");
}
void solve() {
  L[1] = R[1] = 1;
  while (R[sz] <= (long long)2e18) {
    R[sz + 1] = R[sz] * 2 + 1;
    L[sz + 1] = R[sz] + 1;
    sz++;
  }
  int q;
  scanf("%d", &q);
  for (int i = 0, ty; i < q; i++) {
    scanf("%d", &ty);
    if (ty == 1) {
      upd1();
    } else if (ty == 2) {
      upd2();
    } else {
      get();
    }
  }
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  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>
const int inf = (1ll << 30) - 1;
const int maxn = (int)1e5 + 10;
using namespace std;
long long L[111], R[111], SHIFT[111], SHIFTVAL[111];
int sz = 1;
void upd1() {
  long long x, k;
  scanf("%lld%lld", &x, &k);
  int g = 1;
  while (R[g] < x) g++;
  SHIFTVAL[g] -= k;
  SHIFTVAL[g] %= (R[g] - L[g] + 1);
  SHIFTVAL[g] += (R[g] - L[g] + 1);
  SHIFTVAL[g] %= (R[g] - L[g] + 1);
}
void upd2() {
  long long x, k;
  scanf("%lld%lld", &x, &k);
  int g = 1;
  while (R[g] < x) g++;
  while (g <= sz) {
    SHIFTVAL[g] -= k;
    SHIFTVAL[g] %= (R[g] - L[g] + 1);
    SHIFTVAL[g] += (R[g] - L[g] + 1);
    SHIFTVAL[g] %= (R[g] - L[g] + 1);
    k = (k * 2) % (R[sz] - L[sz] + 1);
    g++;
  }
}
void get() {
  long long x;
  scanf("%lld", &x);
  int g = 1;
  while (R[g] < x) g++;
  long long pos = x - L[g];
  pos -= SHIFTVAL[g];
  if (pos < 0) pos += R[g] - L[g] + 1;
  while (g > 0) {
    long long d = (SHIFTVAL[g] + pos) % (R[g] - L[g] + 1);
    printf("%lld", d + L[g]);
    pos /= 2;
    if (g > 1) printf(" ");
    g--;
  }
  printf("\n");
}
void solve() {
  L[1] = R[1] = 1;
  while (R[sz] <= (long long)2e18) {
    R[sz + 1] = R[sz] * 2 + 1;
    L[sz + 1] = R[sz] + 1;
    sz++;
  }
  int q;
  scanf("%d", &q);
  for (int i = 0, ty; i < q; i++) {
    scanf("%d", &ty);
    if (ty == 1) {
      upd1();
    } else if (ty == 2) {
      upd2();
    } else {
      get();
    }
  }
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```