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 double eps = 1e-7;
const int N = 600010;
const int M = 200010;
const int P = 1000000007;
const int Inf = 1000000007;
const int Fni = -1000000007;
inline long long qr() {
  long long r = 0, f = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c <= '9' && c >= '0') {
    r = (r << 1) + (r << 3) + c - '0', c = getchar();
  }
  return r * f;
}
template <typename T>
void debug(T x, char c = '\n') {
  cerr << x << c;
}
template <typename T>
void debug(T *const tp, int minn, int maxn, char v = ' ', char c = '\n') {
  for (int i = minn; i <= maxn; ++i) debug(tp[i], v);
  cerr << c;
}
template <typename T>
inline void chkmin(T &x, T y) {
  x = x > y ? y : x;
}
template <typename T>
inline void chkmax(T &x, T y) {
  x = x < y ? y : x;
}
template <typename T>
inline void add(T &x, long long y, int mod = P) {
  x += y;
  x = x >= mod ? x - mod : x;
}
template <typename T>
inline void dec(T &x, long long y, int mod = P) {
  x -= y;
  x = x < 0 ? x + mod : x;
}
template <typename T>
inline T addn(T x, long long y, int mod = P) {
  x += y;
  return (x = x > mod ? x - mod : x);
}
template <typename T>
inline T decn(T x, long long y, int mod = P) {
  x -= y;
  return (x = x < 0 ? x + mod : x);
}
int T;
map<long long, long long> buc1, buc2;
int main() {
  cin >> T;
  long long x, y, z, t, yf;
  while (T--) {
    x = qr(), y = qr();
    if (x == 1)
      y = log2(y), z = qr(), buc1[y] -= z;
    else if (x == 2) {
      y = log2(y), z = qr();
      buc2[y] += z;
    } else {
      printf("%lld ", y);
      long long p = floor(log2(y));
      z = 1ll << p;
      t = (-buc1[p] % z + z) % z, x = y + t;
      if (x >= (z << 1)) x = x % (z << 1) + z;
      y = x;
      while (p > 0) {
        z = 1ll << p;
        t = (buc2[p] % z + z) % z, x = y + t;
        if (x >= (z << 1)) x = x % (z << 1) + z;
        y = x;
        --p;
        z = 1ll << p;
        y = y >> 1, yf = y;
        t = (buc1[p] % z + z) % z, x = y + t;
        if (x >= (z << 1)) x = x % (z << 1) + z;
        y = x;
        printf("%lld ", y);
        y = yf;
      }
      puts("");
    }
  }
}