Problem Description

As we know, Big Number is always troublesome. But it’s really important in our ACM. And today, your task is to write a program to calculate A mod B.

To make the problem easier, I promise that B will be smaller than 100000.

Is it too hard? No, I work it out in 10 minutes, and my program contains less than 25 lines.

Input

The input contains several test cases. Each test case consists of two positive integers A and B. The length of A will not exceed 1000, and B will be smaller than 100000. Process to the end of file.

Output

For each test case, you have to ouput the result of A mod B.

Sample Input

2 3 12 7 152455856554521 3250

Sample Output

2 5 1521

public BigDecimal remainder(BigDecimal divisor)返回其值为 (this % divisor) 的 BigDecimal。

余数由 this.subtract(this.divideToIntegralValue(divisor).multiply(divisor)) 给出。注意，这不是模操作（结果可以为负）。 参数： divisor - 此 BigDecimal 要除以的值。 返回： this % divisor。

import java.math.BigDecimal;import java.util.Scanner;public class Main {    public static void main(String[] args) {        Scanner sc = new Scanner(System.in);        while(sc.hasNext()){            BigDecimal a = sc.nextBigDecimal();            int b = sc.nextInt();            System.out.println(a.remainder(new BigDecimal(b)));        }    }}