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;
struct P {
  int x1, y1, x2, y2;
};
vector<int> v[111];
int a, c, b, k, n, d, m;
int l[1222];
long long o[3255];
int dx[10] = {0, 1, 0, -1, 1, 1, -1, -1}, dy[10] = {1, 0, -1, 0, -1, 1, -1, 1},
    dz[10] = {0, 0, 0, 0, 1, -1};
int e;
long long x, y, i, mod = 1000000007;
long long z[11];
P j[400];
P u[16];
stack<int> s;
queue<long long> q;
string r1, r;
bool as(P a, P b) { return 0; }
int main() {
  scanf("%d", &a);
  for (int t = 1; t <= a; t++) {
    scanf("%d", &b);
    if (b == 1) {
      long long n, m;
      scanf("%lld %lld", &n, &m);
      c = 1;
      for (i = 1; i <= n; i <<= 1, c++)
        ;
      i >>= 1, c--;
      o[c] += -m + i;
      o[c] %= i;
    } else if (b == 2) {
      long long n, m;
      scanf("%lld %lld", &n, &m);
      c = 1;
      for (i = 1; i <= n; i <<= 1, c++)
        ;
      i >>= 1, c--;
      for (int t = 1; i <= 2000000000000000000; t++, c++, i <<= 1, m <<= 1) {
        o[c] += -m + i;
        o[c] %= i;
      }
    } else {
      long long n;
      scanf("%lld", &n);
      c = 1;
      for (i = 1; i <= n; i <<= 1, c++)
        ;
      i >>= 1, c--;
      n -= o[c];
      if (n < i)
        n += i;
      else if (n >= i * 2)
        n -= i;
      for (; i; i >>= 1, n >>= 1, c--) {
        if (n + o[c] < i)
          printf("%lld ", n + o[c] + i);
        else if (o[c] + n >= i * 2)
          printf("%lld ", o[c] + n - i);
        else
          printf("%lld ", o[c] + n);
      }
      puts("");
    }
  }
}