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Description

There is sequence 1, 12, 123, 1234, …, 12345678910, … . Now you are given two integers A and B, you have to find the number of integers from Ath number to Bth (inclusive) number, which are divisible by 3.

For example, let A = 3. B = 5. So, the numbers in the sequence are, 123, 1234, 12345. And 123, 12345 are divisible by 3. So, the result is 2.

Input

Input starts with an integer T (≤ 10000), denoting the number of test cases.

Each case contains two integers A and B (1 ≤ A ≤ B < 231) in a line.

Output

For each case, print the case number and the total numbers in the sequence between Ath and Bth which are divisible by 3.

Sample Input

2 3 5 10 110 Sample Output Case 1: 2 Case 2: 67

找规律题~~

#include
   
    int main(){    int T,a,b,nl=1;    long long i,ans=0,sum=0;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&a,&b);        a--;        sum=a/3*2;        if(a%3==2)            sum++;        ans=b/3*2;        if(b%3==2)            ans++;            ans-=sum;        printf("Case %d: %lld\n",nl++,ans);    }    return 0;}
   

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