A - Bi-shoe and Phi-shoe

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.

Input

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, 106].

Output

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

Sample Input

3

5

1 2 3 4 5

6

10 11 12 13 14 15

2

1 1

Sample Output

Case 1: 22 Xukha

Case 2: 88 Xukha

Case 3: 4 Xukha

题意：给出n个幸运数字，找到欧拉函数值大于等于这些幸运数字的x，使x的值最小。

思路：因为素数的欧拉函数值就是其减1，这个题就是求最小的大于该幸运数字的素数，然后加起来即可。先进行一个素数打表，然后找到最小的大于幸运数字的素数，相加，最后得出答案。代码如下:

#include<stdio.h>
#include<string.h>
int a[1000010];
int b[1000010];
int main()
{
    int T,n,i,j,t,k=1;
    memset(a,0,sizeof(a));
    for (i = 2; i<=1000010; i++)
    {
        if (a[i]==0)  //0是素数
        {
            for (j=i+i; j<=1000010; j+=i)
                a[j]=1;
        }
    }
    a[1]=1;
    scanf("%d",&T);
    while(T--)
    {
        long long int sum=0;
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
             scanf("%d",&b[i]);
             for(j=b[i]+1;;j++) //此处要加1，不然t-1可能小于b[i];
             {
                 if(a[j]==0)
                 {
                     t=j;
                    break;
                 }
             }
             sum=sum+t;
        }
        printf("Case %d: %lld Xukha\n",k++,sum);
    }
    return 0;
}