Given the value of N, you will have to find the value of S. The definition of S is given in the following code: S=0; for(i=1;i<=N;i++) for(j=1;j<=N;j++) if((N % i)==0 && (N % j)==0) S+=gcd(i,j); /Here ‘gcd()’ is a function that finds the greatest common divisor of the two input numbers. ‘%’ is standard remainder sign from C/C++/java syntax where ‘a % b’ is the remainder of a modulo b, so ‘(N % i)== 0 && (N % j) == 0’meansN isdivisiblebybothiandj/ Input First line of the input is T (T ≤ 100), then T test cases follows in next T lines. Each line contains an integer N (1 ≤ N ≤ 100000000000000 or 1014). The meaning of N is given in the problem statement. Output For each test case print a line in ‘Case I: S’ format where I is case number and S is the value for the N of this case. The value of S will fit in a 64-bit signed integer. Sample Input 12 1 2 3 4 5 6 7 8 9 10 1000 10000 Sample Output Case 1: 1 Case 2: 5 Case 3: 6 Case 4: 15 Case 5: 8 Case 6: 30 Case 7: 10
2/2 Case 8: 37 Case 9: 23 Case 10: 40 Case 11: 8584 Case 12: 97027