Solve the programming task below in a Python markdown code block.
You are given two positive integers N and K. You have to perform the following operation exactly K times:
- For the current value of N, choose any positive integer D such that D is a [divisor] of N and multiply D with N.  
Formally, N := (N * D) such that D is a divisor of current value of N.

Print the sum of all distinct values of the final N you can receive after performing the above operation exactly K times. Since the answer can be large, print it modulo 10^{9} + 7.

------ Input Format ------ 

- The first line of input will contain a single integer T, denoting the number of test cases.
- Each test case contains two space-separated integers N and K respectively, the initial number and the number of operations.

------ Output Format ------ 

For each test case, output on a new line the sum of all distinct values of the final N you can receive after performing the given operation exactly K times. Since the answer can be large, print it modulo 10^{9} + 7.

------ Constraints ------ 

$1 ≤ T ≤ 1000$
$1 ≤ N ≤ 10^{7}$
$1 ≤ K ≤ 10^{5}$

----- Sample Input 1 ------ 
3
1 5
2 2
10 1
----- Sample Output 1 ------ 
1
30
180
----- explanation 1 ------ 
Test case $1$: $1$ is the only divisor of $1$. So, the value remains unchanged after the operations. Thus, there is only one distinct value after $5$ operations, which is $1$.

Test case $2$: 
- Operation $1$: Initially, $N = 2$ has divisors $1$ and $2$. Thus, after $1$ operation, $N$ can be either $2\cdot 1 = 2$ or $2\cdot 2 = 4$.
- Operation $2$: If $N=2$, the divisors are $1$ and $2$ which can lead to the final values as $2\cdot 1 = 2$ and $2\cdot 2 = 4$. If $N = 4$, the divisors are $1, 2, $ and $4$. Thus, the final values can be $4\cdot 1 = 4, 4\cdot 2 = 8, $ and $4\cdot 4 = 16$ .

The distinct values that can be obtained after applying the operation $2$ times on $N = 2$ are $\{2, 4, 8, 16\}$, and $2 + 4 + 8 + 16 = 30$.

Test case $3$: The numbers $10 = 10 \cdot 1$, $20 = 10 \cdot 2$, $50 = 10 \cdot 5$ and $100 = 10 \cdot 10$ can be obtained after applying the operation $1$ time on $N=10$, and $10 + 20 + 50 + 100 = 180$.