Due to COVID19 all employees of chemical factory are quarantined in home. So, company is organized by automated robots. There are $N$ Containers in company, which are labelled with $1$ to $N$ numbers. There are Total $N$ robots in Company, which are labelled $1$ to $N$. 
Each Robot adds $1$ litter Chemical in containers whose labels are multiple of robots’ label. (E.g. robot labelled 2 will add Chemical in containers labelled 2, 4, 6, ….). 
Robots follow few rules mentioned below:
- Chemical with $pH<7$ is acidic.
- Chemical with $pH=7$ is neutral.
- Chemical with $pH>7$ is basic.
- Mixture of $1$ litter acid and $1$ litter base makes neutral solution.
- If Container is empty or pH of Chemical in container is equal to 
- $7$ then robot adds $1$ litter acidic Chemical.
- If pH of Chemical in container is less than $7$ then robots adds $1$ litter basic Chemical.
Now Chemical is packed in $4$ ways:
- Containers with exactly $2$ litter acid and $1$ litter base are packed at $P1$ cost. 
- Containers with acidic Chemical are packed at $P2$ cost. (Here Don’t consider containers with exactly $2$ litter acid and $1$ litter base as we counted them in above rule)
- Containers with $pH=7$ Chemical is packed at $P3$ cost.
- Containers with Basic Chemical are packed at $P4$ cost.
Now $1$ employee given task to find packing cost of containers with labels in range $[ k1, k2]$ (Inclusive). Help him out. As Cost can be large find cost in modulo of $100000007$
-----Input:-----
- The first line of the input contains a single integer T denoting the number of test cases. The description of $T$ test cases follows.
- The first line of each test case contains $3$ space separated Integers $N$, $K1$ and $K2$.
- The second and last line contains $4$ space separated Integers $P1$, $P2$, $P3$ and $P4$
-----Output:-----
For each test case, print total cost of packing of containers modulo 100000007.
-----Constraints-----
- $1 \leq T \leq 50000$
- $1 \leq P1, P2, P3, P4 \leq 2*10^7$
- $2 \leq N \leq 2*10^7$
- $1 \leq K1, K2 \leq N$
-----Subtasks-----Subtask #1 (20 points):
- $1 \leq T \leq 10$
- $1 \leq P1, P2, P3, P4 \leq 2*10^7$
- $2 \leq N \leq 1000$
- $1 \leq K1, K2 \leq N$Subtask #2 (80 points) :
- Original Constraint
-----Sample Input:-----
1
4 1 4
2 2 2 2
-----Sample Output:-----
8
-----Explanation:-----
Let’s use notation, 
0 for acidic Chemical, 
1 for Chemical with pH=7, 
2 for basic Chemical,
3 for Containers with exactly 2 litter acid and 1 litter base,
-1 for empty containers
Initially 4 containers are empty so
[-1, -1, -1, -1]
Robot 1 adds 1 litter acid in all containers.
[0, 0, 0, 0]
Robot 2 adds base in containers labelled 2, 4
[0, 1, 0, 1]
Robot 3 adds base in containers labelled 3
[0, 1, 1, 1]
Robot 4 adds acid in containers labelled 4 because it contains Chemical with pH=7. After adding acid, it also contains exactly 2 litter acid and 1 litter base, so denoting it with 3.
[0, 1, 1, 3]
Corresponding cost of packing container
[P2, P3, P3, P1]
Now, calculate total cost,
Cost = P2+P3+P3+P1 = 2+2+2+2 = 8
primes={}
def SieveOfEratosthenes(n): 
      
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
          
        
        if (prime[p] == True): 
              
            # Update all multiples of p 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
      
    # Print all prime numbers 
    for p in range(2, n+1): 
        if prime[p]: 
            primes[p]=1

        
t=int(input())

SieveOfEratosthenes(15000)
factors={1:2,2:3}

for i in range(3,15000):
    if i in primes:
        factors[i*i]=1
    else:
        factors[i*i]=2

for case in range(t):
    n,k1,k2=list([int(x) for x in input().split()])
    p1,p2,p3,p4=list([int(x) for x in input().split()])
    costs=[0,p1%100000007,p2%100000007,p3%100000007,p4%100000007]
    cost=0
    numbers_to_remove=0
    for number in factors:
        if number >=k1 and number <=k2:
            cost=cost+costs[factors[number]]
            numbers_to_remove+=1
    cost=(cost%100000007+(k2-k1+1-numbers_to_remove)*costs[3]%100000007)%100000007
    
    print(cost%100000007)
            
            

primes={}
def SieveOfEratosthenes(n): 
      
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
          
        
        if (prime[p] == True): 
              
            # Update all multiples of p 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
      
    # Print all prime numbers 
    for p in range(2, n+1): 
        if prime[p]: 
            primes[p]=1

        
t=int(input())

SieveOfEratosthenes(15000)
factors={1:2,2:3}

for i in range(3,15000):
    if i in primes:
        factors[i*i]=1
    else:
        factors[i*i]=2

for case in range(t):
    n,k1,k2=list([int(x) for x in input().split()])
    p1,p2,p3,p4=list([int(x) for x in input().split()])
    costs=[0,p1%100000007,p2%100000007,p3%100000007,p4%100000007]
    cost=0
    numbers_to_remove=0
    for number in factors:
        if number >=k1 and number <=k2:
            cost=cost+costs[factors[number]]
            numbers_to_remove+=1
    cost=(cost%100000007+(k2-k1+1-numbers_to_remove)*costs[3]%100000007)%100000007
    
    print(cost%100000007)
            
            
