Goldbach`s Conjecture

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

Note

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...

The question ： Give several sets of test data , Each group gives one n, ask n A sum that can be divided into pairs of primes .

Ideas ： Progressiveness a prime number makes a table , All primes in the data range are stored in a number , Of course, it has been arranged from small to large , Then from the first in the array 1 To the last convenience , If n Subtracting the value of this element is a prime number num++, If the element is greater than or equal to n/2+1, End convenience . Output num The value of the can . The code is as follows ：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
bool a[10000001];
int prime[666666];
int main()
{
    int i,j,h,n,T,num,k=1,l=0;
    for(i=2; i<=10000000; i++)
    {
        if(a[i]==false)
        {
            prime[l++]=i;
            for(j=i+i; j<=10000000; j=j+i)
                a[j]=true;
        }
    }
    a[0]=a[1]=true;
    scanf("%d",&T);
    while(T--)
    {
        num=0;
        scanf("%d",&n);
        for(i=0; i<l; i++)
        {
            if(prime[i]>=n/2+1)
                break;
            h=n-prime[i];
            if(a[h]==false)
            {
                num++;
            }
        }
        printf("Case %d: %d\n",k++,num);
    }
    return 0;
}