Mathematically some problems look hard. But with the help of the computer, some problems can be easily solvable.

In this problem, you will be given two integers a and b. You have to find the summation of the scores of the numbers from a to b (inclusive). The score of a number is defined as the following function.

score (x) = n², where n is the number of relatively prime numbers with x, which are smaller than x

For example,

For 6, the relatively prime numbers with 6 are 1 and 5. So, score (6) = 2² = 4.

For 8, the relatively prime numbers with 8 are 1, 3, 5 and 7. So, score (8) = 4² = 16.

Now you have to solve this task.

Input

Input starts with an integer T (≤ 10⁵), denoting the number of test cases.

Each case will contain two integers a and b (2 ≤ a ≤ b ≤ 5 * 10⁶).

Output

For each case, print the case number and the summation of all the scores from a to b.

Note

Euler's totient function  applied to a positive integer n is defined to be the number of positive integers less than or equal to n that are relatively prime to n.  is read "phi of n."

Given the general prime factorization of , one can compute  using the formula

思路：就是求欧拉函数，先素数打表，再欧拉函数打表，最后求下前缀和；

 1 #include<stdio.h>

 2 #include<algorithm>

 3 #include<iostream>

 4 #include<string.h>

 5 #include<stdlib.h>

 6 #include<queue>

 7 using namespace std;

 8 bool pr[5*1000006]={0};

 9 int prime[5*100006];

10 unsigned long long  oula[5*1000006];

11 typedef long long LL;

12 int main(void)

13 {

14   LL i,j,p,q;

15   int s,k;

16   scanf("%d",&k);pr[0]=true;

17   pr[1]=true;

18   for(i=0;i<5*1000006;i++)

19     oula[i]=i;

20   for(i=2;i<10000;i++)

21   {

22       if(!pr[i])

23       {

24           for(j=i;i*j<=5*1000000;j++)

25           {

26               pr[i*j]=true;

27           }

28       }

29   }int cnt=0;

30   for(i=2;i<5*1000005;i++)

31   {

32       if(pr[i]==false)

33         prime[cnt++]=i;

34   }

35   for(i=0;i<cnt;i++)

36   {

37       for(j=1;prime[i]*j<=5*1000000;j++)

38       {

39           oula[prime[i]*j]=oula[prime[i]*j]/prime[i]*(prime[i]-1);

40       }

41   }

42   for(i=2;i<=1000000*5;i++)

43   {

44       oula[i]=oula[i-1]+oula[i]*oula[i];

45   }

46   for(s=1;s<=k;s++)

47   {

48       scanf("%lld %lld",&p,&q);printf("Case %d: ",s);

49       printf("%llu\n",oula[q]-oula[p-1]);

50   }

51   return 0;

52 }