Time Limit: 1 second(s) | Memory Limit: 32 MB |
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 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.
For each case of input you have to print the case number and the result according to the pseudo code.
Sample Input |
Output for Sample Input |
3 12 2 12 1 36 2 |
Case 1: 403 Case 2: 28 Case 3: 3751 |
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JANE ALAM JAN |
Copyright © 2012
LightOJ, Jane Alam Jan |