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;
const int maxn = 63;
long long tb[maxn], ad[maxn];
long long qry(int d, long long x) {
  long long beg = tb[d] - (1ll << (d - 1)) + 1;
  long long ans = beg + x - 1 - ad[d];
  if (ans < beg || ans > tb[d]) {
    if (ans < beg) {
      ans = tb[d] - (beg - ans) + 1;
    } else {
      ans = beg + ans - tb[d] - 1;
    }
  }
  return ans;
}
int get(long long x) { return lower_bound(tb + 1, tb + maxn - 1, x) - tb; }
long long getorder(int d, long long x) {
  long long beg = tb[d] - (1ll << (d - 1)) + 1;
  x -= (tb[d] - (1ll << (d - 1)));
  x += ad[d];
  if (x > (1ll << (d - 1))) {
    x -= (1ll << (d - 1));
  } else if (x <= 0) {
    x += (1ll << (d - 1));
  }
  return x;
}
void printpath(long long x) {
  int d = get(x);
  long long order = getorder(d, x);
  for (int i = d; i >= 1; --i) {
    printf("%lld ", x);
    if (i > 1) x = qry(i - 1, order = (order + 1) / 2);
  }
  puts("");
}
void move(int d, long long k) {
  long long len = 1ll << (d - 1);
  k %= len;
  ad[d] += k;
  ad[d] %= len;
}
int main() {
  for (int i = 1; i < maxn; ++i) {
    tb[i] = (1ll << i) - 1;
  }
  int q;
  scanf("%d", &q);
  while (q--) {
    int t;
    long long x, k;
    scanf("%d", &t);
    if (t <= 2) {
      scanf("%lld%lld", &x, &k);
      if (t == 1) {
        move(get(x), k);
      } else {
        int d = get(x);
        k %= 1ll << (d - 1);
        for (int i = d; i < maxn; ++i) {
          move(i, k);
          k *= 2;
        }
      }
    } else {
      scanf("%lld", &x);
      printpath(x);
    }
  }
}