## BASECONV - Base Conversion

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Leo didn't do all the job in his last problem, somebody gave him the numbers in a convenient base. It was the bottleneck of the problem... Now your task is to do this job.

### Input

The first line of input contains three integers T, the number of test cases, B1, the first base, B2, the second base.
For each test case, on the first line your are given one integer k.
On the second line you are given k integers : the digits of N in base B1.

N = a0×B10 + ... + ai×B1i + ... + ak-1×B1k-1

### Output

For each test case, you have to print the number N in base B2. See sample for details.

### Example

```Input:
1 10 100
5
5 4 3 2 1
```
```Output:
3
45 23 1
```

### Explanations

For the lonely case, N = 5×100 + 4×101 + 3×102 + 2×103 + 1×104 = 12345.
We have: N = 45×1000 + 23×1001 + 1×1002. You have to print 3, the number of digits, then the digits: 45, 23 and 1.

### Constraints

```0 < T <= 50
1 < B1,B2 <= 10^9
1 < k <= 10000
0 <= ai < B1  , ak-1>0
```

Time limit is sqrt(T_basic_pike_code * T_awaited_python_code) = sqrt(13.34*6.97), based on my Python3/Pike experiments.
You may try before the tutorial edition.
Have fun ;-)

Edit(2017-02-11) : With compiler updates, a new time limit is set.
Time limit is sqrt(T_basic_pike_code * T_awaited_python_code) = sqrt(3.93*1.57), based on my Python3/Pike experiments.
Thanks @Blue_Mary for pointing this out.