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个数Xi,求满足Φ (ki)>=x的最小的数ki,求所有ki的和，

Φ (n)为欧拉函数,但Φ (1)=0。对于质数p,Φ (p)=p-1.

分析：大于Xi的第一个素数就是最小的ki。

暂时证明不了这个：设素数y1<=x<y2,（则Euler(y1)=y1-1<x,Euler(y2)=y2-1>=x）

那么是否存在整数T(x<T<y2),使Euler(T)>=x

#include
       
        #include
        
         using namespace std; int a[1000005];int p[1000000];int cnt=0;void prime(){    for(int i=4;i<1000005;i+=2) a[i]=1;    p[cnt++]=2;    for(int i=3;i<1000005;i++)    {        if(a[i]==0)        {            p[cnt++]=i;            for(int j=i+i;j<1000005;j+=i)            a[j]=1;        }    }}int main(){    int T,N,cas=0;    prime();    scanf("%d",&T);    while(T--)    {        scanf("%d",&N);        long long ans=0;int x;        while(N--)        {            scanf("%d",&x);            ans+=p[lower_bound(p,p+cnt,x+1)-p];        }        printf("Case %d: %lld Xukha\n",++cas,ans);    }    return 0;}