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 thann which are relatively prime (having no common divisor other than 1) ton. 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 containsn 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

欧拉矩阵的变形，先素数打表，再找离输入数据最近的素数

#include<bits/stdc++.h>using namespace std;int n;long long int a[1000100] = {1, 1, 0};int b[100010];int Eular(long long int n);long long Geta();int main(){    int T, i, j;    long long int sum;    scanf("%d", &T);    Eular(1000010);    for(j = 1; j <= T; j++)    {        scanf("%d", &n);        for(i = 0; i < n; i++)            scanf("%d", &b[i]);        sum = Geta();        printf("Case %d: %lld Xukha\n", j, sum);    }}int Eular(long long int n){    for(int i = 2; i <= n; i++)    {        if(!a[i])        {            for(int j = i + i; j <= 1000010; j += i)                a[j] = 1;        }    }    return 0;}long long int Geta(){    long long int ans = 0;    for(int i = 0; i < n; i++)    {        for(int j = b[i]; j < 1000010; j++)        {            if(!a[j] && j > b[i])            {                  ans += j;                  break;            }        }    }    return ans;}