/*例如10000对m求余：
* 10000%m
* ==(10%m*1000%m)%m
* ==(10%m*(10%m*100%m)%m)%m  
* ==(10%m*(10%m*(10%m*10%m)%m)%m)%m
* 用代码表示就是:
* 假设10000是字符串长度是len 
*/

/*
* 如123对m求余 
* 123%m 
* ==((12%m*10%m)%m+3%m)%m 
* ==(((10%m+2%m)%m*10%m)%m+3%m)%m 
* ==((((1%m*10%m)%m+2%m)%m*10%m)%m+3%m)%m 
*/
gets(str);
int ans=0;
for(i=0;i<len;i++) 
{
	ans=ans*10+str[i];
	ans=ans%m;
}

#include<stdio.h>
#include<string.h>
#define MAX 1100
int main()
{
	int n,m,j,i,s,t;
	char p[MAX];
	while(scanf("%s",p)!=EOF)
	{
		scanf("%d",&n);
		int l=strlen(p);
		s=0;
		for(i=0;i<l;i++)
		{
			s=s*10+p[i]-‘0‘;
			s=s%n;
		}
		printf("%d\n",s);
	}
	return 0;
}