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Goldbach's conjecture is one of the oldest unsolved problems in number theory and in all of mathematics. It states:

Every even integer, greater than 2, can be expressed as the sum of two primes [1].

Now your task is to check whether this conjecture holds for integers up to 10⁷.

Input

Input starts with an integer T (≤ 300), denoting the number of test cases.

Each case starts with a line containing an integer n (4 ≤ n ≤ 10⁷, n is even).

Output

For each case, print the case number and the number of ways you can express n as sum of two primes. To be more specific, we want to find the number of(a, b) where

1)      Both a and b are prime

2)      a + b = n

3)      a ≤ b

Sample Input

2

6

4

Sample Output

Case 1: 1

Case 2: 1

Hint

1.      An integer is said to be prime, if it is divisible by exactly two different integers. First few primes are 2, 3, 5, 7, 11, 13, ...

数据太大会超限，C/C++中 bool类型占用一个字节，int类型占用4个字节，在素数打表中采用bool类型可以节约不少内存。

把筛得的素数存起来，后来找时更方便

#include <iostream>#include <cstdio>#include <cstring>#include <vector>using namespace std;bool a[10000005];int p[1000000];void prime(){    memset(a,false,sizeof(a));    memset(p,0,sizeof(p));    int i,j;    int k=0;    for(i=2;i<10000000;i++)    {        if(!a[i])        {          p[k++]=i;          for(j=2*i;j<10000000;j+=i)            a[j]=true;        }    }}int main(){  int t,n;   prime();  scanf("%d",&t);  int r=1;  while(t--)  {      int sum=0;      scanf("%d",&n);      for(int i=0;p[i]<=n/2;i++)      {         if(!a[n-p[i]])            sum++;      }      printf("Case %d: %d\n",r++,sum);  }}