First Common Multiple at or Above L

Source: Phước Hưng OJ

Version: Phuoc Hung OJ Extended

Problem Statement

Given three positive integers $a$, $b$, and $L$, find the smallest positive integer $x$ such that $x\ge L$ and $x$ is divisible by both $a$ and $b$. If $L$ is already divisible by both, the answer is $L$.

Input

One line contains three positive integers $a$, $b$, and $L$, in that order.

Output

Print one integer: the smallest $x\ge L$ divisible by both $a$ and $b$.

Subtasks

• Subtask 1 (20%): $1\\le a,b\\le10,\\ 1\\le L\\le100$.
• Subtask 2 (30%): $1\\le a,b\\le50,\\ 1\\le L\\le1000$.
• Subtask 3 (50%): $1\\le a,b\\le100,\\ 1\\le L\\le10000$.

Examples

Example 1

Input:

4 6 13

Output:

24

Explanation: The first multiples of 4 at or above 13 are 16, 20 and 24. Only 24 among them is also divisible by 6; no smaller valid number exists.

Example 2

Input:

5 5 10

Output:

10

Explanation: Here $a=b=5$ and $10$ is divisible by both, so the starting bound is the answer.

Example 3

Input:

7 9 1

Output:

63

Explanation: The least positive common multiple of 7 and 9 is 63; none of 1 through 62 works.