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

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

(Xzhilans are really fond of number theory). For yourinformation,Φ (n) = numbers less thann which arerelatively prime (having no common divisor other than 1) ton. So,scoreof 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 eachstudent. 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 scoregreater than or equal to his/her lucky number. Bi-shoe wants to minimize thetotal amount of money spent for buying the bamboos. One unit of bamboo costs 1Xukha. 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 linecontainsn space separated integers denoting the lucky numbers for thestudents. Each lucky number will lie in the range[1,10⁶].

Output

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

题意是知道一串数字。然后规定一个规则，score 为 length以内与length互质的数的个数，每个数字依次分析，对于每个数字，求不小于这个数的score对应的length，然后加起来。

这个题可以转化一下题意，我们知道对于一个length为质数的数，他的score一定等于length - 1，然后对于任意一个合数，他的score一定小于它前面的那个质数的score，所以题意就转化为求解比当前数字大的最小的质数的大小，将所有的累计，得到最后的答案。

欧拉筛法解决。

代码如下：

/*************************************************************************
	> File Name: Bi-shoe_and_Phi-shoe.cpp
	> Author: Zhanghaoran
	> Mail: chilumanxi@xiyoulinux.org
	> Created Time: Tue 08 Dec 2015 12:38:03 PM CST
 ************************************************************************/

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cstdlib>

using namespace std;
const int Max = 1000100;
int Prim[Max];
int Isprim[Max];

void euler(){
    int ans = 0;
    Isprim[0] = Isprim[1] = 1;
    for(int i = 2; i <= Max - 1; i ++){
        if(!Isprim[i])
            Prim[ans ++] = i;
        for(int j = 0; j < ans &&  Prim[j] * i <= Max - 1; j ++){
            Isprim[Prim[j] * i] = 1;
            if(i % Prim[j] == 0)
                break;
        }
    }
}

int T;
int N;
int main(void){
    cin >> T;
    long sum = 0;
    euler();
    int cas = 0;
    while(T --){
        cin >> N;
        cas ++;
        int temp;
        for(int i = 0; i < N; i ++){
            cin >> temp;
            for(int j = temp + 1; j <= Max - 1; j ++){
                if(!Isprim[j]){
                    sum += j;
                    break;
                }
            }
        }
        printf("Case %d: %ld Xukhan",cas,sum);
        sum = 0;
    }
}