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.

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.

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

  
  

   
   2 3
12 7
152455856554521 3250
  
  

  
  

   
   2
5
1521
  
  
  
  

   
   模板：
  
  
  
  
 
  
  
  
  

   
   
for(i=0;i<len;i++)
{
sum=(sum*10+(s[i]-'0')%m)%m;
}
  
  
  
  

   
   代码：
  
  
  
  
 
   
   
#include<stdio.h>
#include<string.h>
int main()
{
	int i,m;
	char s[1000];
	while(scanf("%s%d",s,&m)!=EOF)
	{
		int sum=0;
		int len=strlen(s);
		for(i=0;i<len;i++)
		 {
		 	sum=(sum*10+(s[i]-'0')%m)%m;
		 }
		 printf("%dn",sum);
    } 
	 
}