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 = 611090;
const long long MOD = 1000000007LL;
const int fx[4][2] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
double pi = acos(-1.0);
inline int lowbit(int x) { return x & (-x); }
void gcd(long long a, long long b, long long &d, long long &x, long long &y) {
  if (!b) {
    d = a;
    x = 1;
    y = 0;
  } else {
    gcd(b, a % b, d, y, x);
    y -= x * (a / b);
  }
}
long long inv(long long a, long long n) {
  long long d, x, y;
  gcd(a, n, d, x, y);
  return (x + n) % n;
}
long long a[70];
int n;
int main() {
  int cases = 1;
  double ans = 0;
  for (int iii = 1; iii <= cases; ++iii) {
    scanf("%d", &n);
    for (int nn = 0; nn < n; ++nn) {
      int op;
      scanf("%d", &op);
      if (op == 1) {
        long long x, y;
        scanf("%lld%lld", &x, &y);
        if (!y) continue;
        long long temp = 61, temp1 = temp;
        while ((1LL << temp) > x) --temp;
        temp1 = temp;
        temp = 1LL << temp;
        if (y < 0) y = ((-y) / temp * temp + y);
        y = (y + (temp)) % (temp);
        a[temp1] = (a[temp1] + y) % temp;
      } else if (op == 2) {
        long long x, y;
        scanf("%lld%lld", &x, &y);
        if (!y) continue;
        long long temp = 61, temp1 = temp;
        while ((1LL << temp) > x) --temp;
        temp1 = temp;
        temp = 1LL << temp;
        if (y < 0) y = ((-y) / temp * temp + y);
        y = (y + (temp)) % (temp);
        for (; temp1 < 61LL; ++temp1) {
          a[temp1] = (a[temp1] + y) % (1LL << temp1);
          y = y * 2 % (2LL << temp1);
        }
      } else {
        long long x;
        scanf("%lld", &x);
        long long temp = 61, temp1 = temp;
        while ((1LL << temp) > x) --temp;
        x = x + a[temp];
        if (x >= (2LL << temp)) x -= (1LL << (temp));
        while (x) {
          printf("%lld",
                 (x + (1LL << temp) - a[temp]) % (1LL << temp) + (1LL << temp));
          x >>= 1;
          --temp;
          if (x)
            printf(" ");
          else
            printf("\n");
        }
      }
    }
  }
}