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 N = 105;
long long delta[N], c[N], l[N], r[N];
long long la, now, x, k;
int cnt = 0;
int q, op;
void out(long long a) {
  if (a >= 10) out(a / 10);
  putchar(a % 10 + '0');
}
int calc(long long x) {
  int ret = 0;
  for (; x; x >>= 1ll) ++ret;
  return ret;
}
int main() {
  now = 0;
  la = now + 1;
  for (int i(0); i <= (62); ++i) {
    now += 1ll << i;
    l[++cnt] = la, r[cnt] = now;
    c[cnt] = 1ll << i;
    la = now + 1;
  }
  scanf("%d", &q);
  while (q--) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%lld%lld", &x, &k);
      int et = calc(x);
      k %= c[et];
      k += c[et];
      k %= c[et];
      delta[et] += k;
      delta[et] %= c[et];
    } else if (op == 2) {
      scanf("%lld%lld", &x, &k);
      int et = calc(x);
      k %= c[et];
      k += c[et];
      k %= c[et];
      for (int i(et); i <= (61); ++i) {
        k %= c[i];
        delta[i] += k;
        delta[i] %= c[i];
        k = k * 2ll;
      }
    } else {
      scanf("%lld", &x);
      out(x);
      int et = calc(x);
      long long pos = x - l[et];
      pos = pos + delta[et];
      pos %= c[et];
      for (int i(et - 1); i >= (1); --i) {
        pos >>= 1ll;
        long long re = pos;
        re = re - delta[i];
        re = re + c[i];
        re %= c[i];
        re += l[i];
        putchar(32);
        out(re);
      }
      putchar(10);
    }
  }
  return 0;
}