Given two integers, a and b, you should check whether a is divisible by b or not. We know that an integer a is divisible by an integer b if and only if there exists an integer c such that a = b * c.

Input

Input starts with an integer T (≤ 525), denoting the number of test cases.

Each case starts with a line containing two integers a (-10²⁰⁰ ≤ a ≤ 10²⁰⁰) and b (|b| > 0, b fits into a 32 bit signed integer). Numbers will not contain leading zeroes.

Output

For each case, print the case number first. Then print ‘divisible‘ if a is divisible by b. Otherwise print ‘not divisible‘.

sum忘记初始化错了好久，还是同余定理

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
#define INF 0x3f3f3f
#define LL long long 
#define MAX 2000002
using namespace std;
int main()
{
	int t,n,j,i;
	int k=1;
	char s[MAX];
	scanf("%d",&t);
	while(t--)
	{
		memset(s,0,sizeof(s));
		scanf("%s%d",s,&n);
		printf("Case %d: ",k++);
		if(n<0)
		    n=-n;
		int len=strlen(s);
		LL sum=0; 
		for(j=0;j<len;j++)
		{	
		    if(s[j]==‘-‘)
			continue;		
			sum=sum*10+(s[j]-‘0‘);
			sum=sum%n;
		}
		if(sum==0)
		printf("divisible\n");
		else
		printf("not divisible\n");
	}
	return 0;
}