Given a triplet of integers (X , Y , Z), such that X ≤ Y and Y ≥ Z, we define f(X , Y , Z) to be (X + Y) * (Y + Z). If either X > Y or Y < Z, or both, then f(X , Y , Z) is defined to be 0.
You are provided three arrays A , B and C of any length (their lengths may or may not be equal).
Your task is to find the sum of f(X , Y , Z) over all triplets (X, Y , Z) where X, Y and Z belong to A, B and C respectively.
Output your sum for each test case modulo 1000000007.

-----Input-----
- The first line contains a single integer, T, which is the number of test cases. The description of each testcase follows:
- The first line of each testcase contains 3 integers: p, q and r. These denote the lengths of A,B and C respectively.
- The second line contains p integers, which are the elements of A
- The third line contains q integers, which are the elements of B
- The fourth line contains r integers, which are the elements of C

-----Output-----
Output the required sum modulo 1000000007 for each test case in a new line.

-----Constraints-----
- 1 ≤ T ≤ 10
- 1 ≤ p, q, r ≤ 100000
- 1 ≤ every array element ≤ 1000000000

-----Subtasks -----
- Subtask #1 (30 points): 1 ≤ p,q,r ≤ 100
- Subtask #2 (70 points): 1 ≤ p,q,r ≤ 100000

-----Example : -----
Input:
1
3 1 3
1 2 3
5
4 5 6

Output:
399

-----Explanation: -----
As there is only one choice for Y which equals to 5, to get a non-zero function value,we can choose any element for X from the set { 1 , 2 , 3 } and for Z from the set { 4 , 5 }
So triplets which give non-zero function values are:
{ 1 , 5 , 4 } : ( 1 + 5 ) * ( 5 + 4 ) = 54
{ 1 , 5 , 5 } : ( 1 + 5 ) * ( 5 + 5 ) = 60
{ 2 , 5 , 4 } : ( 2 + 5 ) * ( 5 + 4 ) = 63
{ 2 , 5 , 5 } : ( 2 + 5 ) * ( 5 + 5 ) = 70
{ 3 , 5 , 4 } : ( 3 + 5 ) * ( 5 + 4 ) = 72
{ 3 , 5 , 5 } : ( 3 + 5 ) * ( 5 + 5 ) = 80
Final answer : 54 + 60 + 63 + 70 + 72 + 80 = 399

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