POWER SUMMATION :')

Author: Nazmul Alam

Leeana Learned Few New Things Few Days Ago , Like:

  1. Find The Summation Of Divisors With Prime Factorization
  2. Modular Arithmetic


So Now Her Uncle Gave Her A Task.

Task Is: You Will Be Given A Number(N) And Another Number(K). Now You Have To Find  Kth Power Summation Of Divisors.

Summation Of Divisors Will Be Huge,  So You Need To Print The Summation Module (M=1000000007).

Like: Divisors Of 6 is: ( 1 2 3 6 ) And K = 2.

so, summation is: 1^K+2^K+3^K+6^K = 1^2 + 2^2 + 3^2 + 6^2= 1+4+9+36 = 50%1000000007=50

Leeana Thinks That You Are A Great Programmer, So She Needs Your Help. Can You Help Her??? :D :D :D 

Input Format

Input Starts With An Integer 1<=T<=500, Denoting The Number Of Test Cases. Each Case Contains An Integer The Number 1<=N<=10^15 and An Integer 1<=K<=10^5 Denoting The Power Of Divisors.

Output Format

For Each Case, Print The Case Number And Kth Power Summation  Of Divisors Of N Module 1000000007. After Each Case Print A New Line. See Sample Input And Output For Better Explanation. :)

Samples

Input
4 6 2 6 1 6 4 6 3
Output
Case 1: 50 Case 2: 12 Case 3: 1394 Case 4: 252

Explanation: case 1: 1^2+2^2+3^2+6^2=50%1000000007=50 case 2: 1^1+2^1+3^1+6^1=12%1000000007=12 case 3: 1^4+2^4+3^4+6^4=1394%1000000007=1394 case 4: 1^3+2^3+3^3+6^3=252%1000000007=252

Limits
Language Time Memory
GNU C 11 2s 1024MB
GNU C++ 14 2s 1024MB
GNU C++ 11 1s 512MB
Statistics
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