#### Given an array βARRβ and an integer βKβ, your task is to find all the count of all sub-arrays whose sum is divisible by the given integer βKβ.

##### Note:

```
If there exists no subarray in the given sequence whose sum is divisible by βKβ then simply return β0β.
```

##### Example:

```
Suppose the given array is βARRβ = { 5, 0, 2, 3, 1} and βK = 5β.
As we can see in below image, there are a total of 6 subarrays that have the total sum divisible by βKβ
So we return the integer 6.
```

```
The first line of input contains an integer βTβ denoting the number of test cases.
The next β2*Tβ lines represent the βTβ test cases.
The first line of each test case contains two space-separated integers the first integer βNβ will denote the number of elements in the array and the second integer denotes integer βKβ.
The second line of each test case contains βNβ space-separated integer that is elements of the array.
```

```
For each test case, print an integer that is the count of all subarray that sum is divisible by βKβ.
```

##### Note:

```
You are not required to print the output explicitly, it has already been taken care of. Just implement the function.
```

##### Constraints:

```
1 <= T <= 50
1 <= K,N <= 10^4
-10^9 <= ARR[i] <= 10^9
Time limit: 1 second
```

##### Sample Input 1:

```
2
3 2
2 3 1
4 1
1 2 3 4
```

##### Sample Output 1:

```
3
10
```

##### Explanation of sample input 1:

```
Test Case 1:
Given βARRβ is { 2, 3,1 } and βKβ is β2β.
All the sub-array with sum is divided by βKβ are -
{ 2 } because the sum is 2 and sum 2 is divisible by 2
{ 3, 1 } because the sum is 3 + 1 = 4 and sum 4 is divisible by 2.
{ 2, 3, 1 } because the sum is 2 + 3 + 1 = 6 and sum 6 is divisible by 2.
Hence there is a total of three subarrays that has sum divisible by 2.
Test Case 2:
Given βARRβ is { 1, 2, 3, 4 } and βKβ is β1β.
Given βKβ is 1 thatβs why each and every sub-arrays sum will be divisible by β1β and with the size of β4β array total number of subarray possible is β( 4*5 /2 ) = 20/2 = 10β.
All possible subarray -
{ 1 }, { 2 }, { 3 }, { 4 }, { 1, 2 }, { 2, 3 }, { 3, 4 }, { 1, 2, 3 }, { 2, 3, 4 }, { 1, 2, 3, 4 } and all subarray sum is divisible by β1β.
Hence there are overall 10 subarrays that has sum divisible by β1β.
```

##### Sample Input 2:

```
2
4 3
1 4 5 2
3 2
1 1 2
```

##### Sample Output 2:

```
2
3
```