A

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：t个测试样例，每个样例 n 个数，对于给定的每个数a[i]，要求一个数x，满足x的欧拉函数值大于a[i]

要找n个满足条件的数x,并且要求n 个数x 的和 cnt 最小，输出 cnt

题解：一个素数 p 的欧拉函数值等于 p-1；所以要找满足条件：欧拉函数值大于a[i] 的最小x；只要找大于a[i] 的最小素数即可，最后输出n个素数的和

#include<iostream>
#include<algorithm>
#include<math.h>
#include<string.h>
#include<string>
#define ll long long
#define mx 1000010
using namespace std;
int prim[mx];
void init()//素数打表
{
    memset(prim,0,sizeof(prim));
    prim[1]=1;
    for(int i=2;i<mx;i++)
    {
        if(prim[i])//是偶数
            continue;
       
        for(int j=i<<1;j<mx;j=j+i)//把素数的倍数标记
            prim[j]=1;
    }
}
int main()
{
    int t,n,x;
    cin>>t;
    init();
    for(int i=1;i<=t;i++)
    {
        cin>>n;
        ll cnt=0;
        for(int j=0;j<n;j++)
        {
            cin>>x;
            for(int k=x+1;k<mx;k++)
            {
                if(prim[k]==0)
                {
                    cnt=cnt+k;
                    break;
                }
            }
        }
        cout<<"Case "<<i<<": "<<cnt<<" Xukha"<<endl;
    }
    return 0;
}