Algorithm title: Codeforces 236B

Link:

Http://www.codeforces.com/problemset/problem/236/B

Topic:

B. Easy Number Challenge

Time limit per test

2 seconds

Memory limit per test

256 Megabytes

Input

Standard input

Output

Standard output

Let ' s denote D (n) as the number of divisors of a positive integer n. You are given three integers a, B and C. Your task is to calculate the following sum:

Find the sum modulo 1073741824 (230).

Input

The contains three space-separated integers a, B and C (1≤a,b,c≤100).

Output

Print a single integer-the required sum modulo 1073741824 (230).

Sample Test (s)

Input

2 2 2

Output

20

Input

5 6 7

Output

1520

Note

For the the example.

D (1 1 1) =d (1) = 1;
D (1 1 2) =d (2) = 2;
D (1 2 1) =d (2) = 2;
D (1 2 2) =d (4) = 3;
D (2 1 1) =d (2) = 2;
D (2 1 2) =d (4) = 3;
D (2 2 1) =d (4) = 3;
D (2 2 2) =d (8) =4.

So this is 1+2+2+3+2+3+3+4=20.

Analysis and Summary:

Bare factor number, data comparison water can be violent, but as long as give a 100 100 100, will be tle by others hack off.

For me this kind of maths weak force, when playing CF can only temporary Baidu a more efficient way ...

Condition: Given any one positive integer n

Requirements: Number of factors to be obtained

Firstly, the conclusion is given that for any integral type N, the decomposition decomposition gets n= p1^x1 * p2^x2* ... * PN^XN;

The number of factors M for n is m= (x1+1) * (x2+1) * ... * (xn+1);

Certification process:

Let's give you an example.

24 = 2^3 * 3^1;

The mass factors are as follows: 2 and 3 indices are 3 and 1.

Then for 2 there are 0 1 2 34 index options, for 3 0 of the index options

So that's 4 * 2 = 8 number of factors

If you still don't understand, then let's list it.

2 3

2^0*3^0=1 2^0*3^1=3

2^1*3^0=2 2^1*3^1=6

2^2*3^0=4 2^2*3^1=12

2^3*3^0=8 2^3*3^1=24