A

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

351 2 3 4 5610 11 12 13 14 1521 1

Sample Output

Case 1: 22 XukhaCase 2: 88 Xukha

找到大于其本身的最小的素数，求和

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>using namespace std;bool p[1234567];void isprime()//素数筛{   memset(p, true, sizeof(p));//初始化   p[1] = false;//1不是素数   for(int i=2;i<1234567;i++)   {      if(p[i])      {         for(int j=2*i;j<1234567;j+=i)//对于其倍数进行同样的操作         {            p[j] = false;         }      }   }}int main(){  int T;  isprime();  scanf("%d", &T);  for(int t=1;t<=T;t++)  {     int n, m;     scanf("%d", &n);      long long int k = 0;//求和，注意使用long long      for(int i=1;i<=n;i++)     {       scanf("%d", &m);       for(int j=m+1;j<1234567;j++)       {         if(p[j])         {           k += j;           break;         }       }     }     printf("Case %d: %lld Xukha\n", t, k);  }  return 0;}