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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, ...

题意：给你一个数k  找出有多少个素数对（素数a+素数b=k  ）a可以等于b   

#include<cstdio>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#include<map>#include<vector>#include<queue>#define  LL  long long//const LL INF=1e18;const int INF=1e7+5;using namespace std;int su[1000000],cnt,n,k;bool a[INF];void aa(){    int cnt=1;    memset(a,1,sizeof(a));    a[0]=a[1]=0;    for(int i=2;i<INF;i++)    {        if(a[i])            su[cnt++]=i;        for(int j=1;j<cnt&&su[j]*i<INF;j++)            a[su[j]*i]=0;    }}int main(){    aa();    int o=1;    int ans;    scanf("%d",&n);    while(n--)    {        ans=0;        scanf("%d",&k);        int k1=k/2;        for(int i=1;su[i]<=k1;i++)        {            if(a[k-su[i]])                ans++;        }        printf("Case %d: %d\n",o++,ans);    }}