G

If we sum up every digit of a number and the result can be exactly divided by 1010 , we say this number is a good number.

You are required to count the number of good numbers in the range from AA to BB , inclusive.

Input

The first line has a number TT (T≤10000T≤10000 ) , indicating the number of test cases.

Each test case comes with a single line with two numbers AA and BB (0≤A≤B≤10180≤A≤B≤1018 ).

Output

For test case XX , output Case #X: first, then output the number of good numbers in a single line.

Sample Input

2
1 10
1 20

Sample Output

Case #1: 0
Case #2: 1

Hint

The answer maybe very large, we recommend you to use long long instead of int.

分析：题目中的数据很大，所以可以往找规律的方面考虑

打表之后会发现，每10个数都会有一个符合要求的数

如果第十个数（整除10的数）是符合要求的数 就再加一个

代码如下：

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
using namespace std;
typedef long long ll;
#define INF 0x7fffffff
int main()
{
    ll a,b,cnta,cntb,h,num,t;
    cin>>t;
    for(ll r=1;r<=t;r++)
    {
    cin>>a>>b;
        a=a-1;
        cnta=0;
        cntb=0;
        cnta=a/10;
        h=a-a%10;
        num=0;
        while(h>0)
        {
            num+=h%10;
            h=h/10;
        }
        if(num%10==0)
        cnta++;
        for(ll i=a-a%10+1;i<=a;i++)
        {
        //    puts("1");
            h=i;
            num=0;
                while(h>0)
        {
            num+=h%10;
            h=h/10;
        }
             if(num%10==0)
               cnta++;
        }
        if(a<0)
        cnta=0;
        cntb=b/10;
            h=b-b%10;
        num=0;
        while(h>0)
        {
            num+=h%10;
            h=h/10;
        }
        if(num%10==0)
        cntb++;
        for(ll i=b-b%10+1;i<=b;i++)
        {
        
            h=i;
            num=0;
                while(h>0)
        {
            num+=h%10;
            h=h/10;
        }
             if(num%10==0)
               cntb++;
        }
    //    cout<<cnta<<endl;
    //    cout<<cntb<<endl;
        printf("Case #%lld: %lld
",r,cntb-cnta);
    
}
}