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 long long mod = 1000000007;
const double eps = 1e-9;
const int inf = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
long long powmod(long long a, long long b) {
  long long res = 1;
  a %= mod;
  assert(b >= 0);
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mod;
    a = a * a % mod;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline int read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (ch < '0' || '9' < ch) f |= ch == '-', ch = getchar();
  while ('0' <= ch && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return f ? -x : x;
}
long long p1[70], p2[70];
void do1(long long x, long long k) {
  int lev = 0;
  while ((x >> lev) > 1) lev++;
  p1[lev] += k;
}
void do2(long long x, long long k) {
  int lev = 0;
  while ((x >> lev) > 1) lev++;
  p2[lev] += k;
}
void print(long long x) {
  int lev = 0;
  while ((x >> lev) > 1) lev++;
  long long mo = (1LL << lev);
  printf("%I64d", x);
  x = (x + p1[lev]) % mo;
  while (x < mo) x += mo;
  x = (x + p2[lev]) % mo;
  while (x < mo) x += mo;
  while (lev) {
    x >>= 1;
    lev--;
    mo = (1LL << lev);
    long long y = (x - p1[lev]) % mo;
    while (y < mo) y += mo;
    x = (x + p2[lev]) % mo;
    while (x < mo) x += mo;
    printf(" %I64d", y);
  }
  puts("");
}
int main() {
  memset(p1, 0, sizeof(p1));
  memset(p2, 0, sizeof(p2));
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    long long op, x, k;
    scanf("%lld", &op);
    if (op == 1) {
      scanf("%lld%lld", &x, &k);
      do1(x, k);
    } else if (op == 2) {
      scanf("%lld%lld", &x, &k);
      do2(x, k);
    } else {
      scanf("%lld", &x);
      print(x);
    }
  }
  return 0;
}