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.

?

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i,a,b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i,n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a,x ) memset ( a,x,sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int INF = 0x3f3f3f3f;
const int maxn=1e6+5;
int phi[maxn+10];
int num[maxn];
int t,n;
void init()
{
    REPF(i,maxn)  phi[i]=0;
    phi[1]=1;
    REPF(i,maxn)
      if(!phi[i])
        for(int j=i;j<=maxn;j+=i)
        {
            if(!phi[j])  phi[j]=j;
            phi[j]=phi[j]/i*(i-1);
        }
}
int main()
{
    init();
    scanf("%d",&t);
    int cas=1;
    while(t--)
    {
        scanf("%d",&n);
        REP(i,n)  scanf("%d",&num[i]);
        sort(num,num+n);
        LL sum=0;
        for(int i=0,j=2;i<n;)
        {
            if(phi[j]>=num[i])
            {
                sum+=j;
                i++;
            }
            else j++;
        }
        printf("Case %d: %lld Xukhan",cas++,sum);
    }
    return 0;
}