題目表述：

Bamboo Pole-vault is a massively popular sport in Xzhiland. And Master Phi-shoe is a very popular coach for his success. He needs some bamboos for his students, so he asked his assistant Bi-Shoe to go to the market and buy them. Plenty of Bamboos of all possible integer lengths (yes!) are available in the market. According to Xzhila tradition,

Score of a bamboo = Φ (bamboo's length)

(Xzhilans are really fond of number theory). For your information, Φ (n) = numbers less than n which are relatively prime (having no common divisor other than 1) to n. So, score of a bamboo of length 9 is 6 as 1, 2, 4, 5, 7, 8 are relatively prime to 9.

The assistant Bi-shoe has to buy one bamboo for each student. As a twist, each pole-vault student of Phi-shoe has a lucky number. Bi-shoe wants to buy bamboos such that each of them gets a bamboo with a score greater than or equal to his/her lucky number. Bi-shoe wants to minimize the total amount of money spent for buying the bamboos. One unit of bamboo costs 1 Xukha. Help him.

輸入：

Input starts with an integer T (≤ 100), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 10000) denoting the number of students of Phi-shoe. The next line contains n space separated integers denoting the lucky numbers for the students. Each lucky number will lie in the range [1, 10的6次方]

輸出：

For each case, print the case number and the minimum possible money spent for buying the bamboos. See the samples for details.

樣例輸入：

3

5

1 2 3 4 5

6

10 11 12 13 14 15

2

1 1

樣例輸出：

Case 1: 22 Xukha

Case 2: 88 Xukha

Case 3: 4 Xukha

這個題的意思是給出幾個幸運數字，對於每一個幸運數字要求出一個N，並且N的尤拉函式的值不小於這個幸運數字，並且所有N的和要最小，所以每個N最小那麼N的和就是最小。對於每一個幸運數字，他的N的最小值就是大於幸運數字的第一個素數。在範圍內，把所有的素數都標記為0.

程式碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
#define maxn 1000010
int isprime[maxn];
void get_prime()//篩選素數，過程可以用筆寫一下，便於理解
{
    int i,j;
    memset(isprime,0,sizeof(isprime));
    for(i=2;i<1001;i++)
    {
        if(isprime[i]==0)
        {
            for(j=i*i;j<maxn;j+=i)
            {
                isprime[j]=1;
            }
        }
    }
}
int main()
{
    int t,n;
    int k;
    int ans=1;
    scanf("%d",&t);
    get_prime();
    while(t--)
    {
        long long sum=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            scanf("%d",&k);
            for(int j=k+1;j<=maxn;j++)
            {
                if(isprime[j]==0)
                {
                     sum+=j;
                     break;
                }

            }
        }
        printf("Case %d: %lld Xukha\n",ans++,sum);
    }
    return 0;
}