F - Efficient Pseudo Code

Sometimes it's quite useful to write pseudo codes for problems. Actually you can write the necessary steps to solve a particular problem. In this problem you are given a pseudo code to solve a problem and you have to implement the pseudo code efficiently. Simple! Isn't it? 😃

pseudo code
{
    take two integers n and m
    let p = n ^ m (n to the power m)
    let sum = summation of all the divisors of p
    let result = sum MODULO 1000,000,007
}

Now given n and m you have to find the desired result from the pseudo code. For example if n = 12 and m = 2. Then if we follow the pseudo code, we get

pseudo code
{
    take two integers n and m
    so, n = 12 and m = 2
    let p = n ^ m (n to the power m)
    so, p = 144
    let sum = summation of all the divisors of p
    so, sum = 403, since the divisors of p are 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144
    let result = sum MODULO 1000,000,007
    so, result = 403
}

Input

Input starts with an integer T (≤ 5000), denoting the number of test cases.
Each test case will contain two integers, n (1 ≤ n) and m (0 ≤ m). Each of n and m will be fit into a 32 bit signed integer.

Output

For each case of input you have to print the case number and the result according to the pseudo code.

Sample Input

3
12 2
12 1
36 2

Sample Output

Case 1: 403
Case 2: 28
Case 3: 3751