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 int lvl[64] = {0};
long long int sft[64] = {0};
void init() {
  long long int i, j, k = 1;
  for (i = 1; i < 64; i++) {
    lvl[i] = k;
    k *= 2;
  }
  return;
}
long long int levelof(long long int n) {
  long long int i, j, k, l;
  k = 1;
  l = 0;
  for (i = 1; i < 64; i++) {
    l += lvl[i];
    if (l >= n) {
      return k;
    }
    k++;
  }
  return k;
}
void addshift(long long int lvlofx, long long int n) {
  long long int i, j, k, m;
  if (n == 0) {
    return;
  }
  if (lvlofx == 1) {
    return;
  }
  n %= lvl[lvlofx];
  if (n < 0) {
    n += lvl[lvlofx];
  }
  sft[lvlofx] += n;
  sft[lvlofx] %= lvl[lvlofx];
  return;
}
void addlayershift(long long int lvlofx, long long int n) {
  long long int i, j, k, l, m;
  if (n == 0) {
    return;
  }
  if (lvlofx == 1) {
    return;
  }
  for (i = lvlofx; i < 64; i++) {
    n %= lvl[i];
    addshift(i, n);
    n *= 2;
  }
  return;
}
void print(long long int n) {
  long long int i, j, k, l;
  printf("%lld ", n);
  long long int lvlofx = levelof(n);
  k = n + sft[lvlofx];
  if (k >= lvl[lvlofx + 1]) {
    l = k - lvl[lvlofx + 1];
    k = l + lvl[lvlofx];
  }
  for (i = lvlofx - 1; i > 0; i--) {
    l = k / 2;
    j = l - sft[i];
    if (j < lvl[i]) {
      j += lvl[i];
    }
    printf("%lld ", j);
    k /= 2;
  }
  printf("\n");
  return;
}
int main() {
  long long int n, q, m, i, j, k, l, chk, x;
  init();
  long long int levelofx;
  scanf("%lld", &q);
  while (q--) {
    scanf("%lld", &chk);
    if (chk == 1) {
      scanf("%lld %lld", &x, &k);
      levelofx = levelof(x);
      addshift(levelofx, k);
    } else if (chk == 2) {
      scanf("%lld %lld", &x, &k);
      levelofx = levelof(x);
      addlayershift(levelofx, k);
    } else {
      scanf("%lld", &x);
      print(x);
    }
  }
}