Description

If a and d are relatively prime positive integers, the arithmetic sequence beginning with a and increasing by d, i.e., a, a + d, a + 2d, a + 3d, a + 4d, ..., contains infinitely many prime numbers. This fact is known as Dirichlet's Theorem on Arithmetic Progressions, which had been conjectured by Johann Carl Friedrich Gauss (1777 - 1855) and was proved by Johann Peter Gustav Lejeune Dirichlet (1805 - 1859) in 1837.

For example, the arithmetic sequence beginning with 2 and increasing by 3, i.e.,

2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, ... ,

contains infinitely many prime numbers

2, 5, 11, 17, 23, 29, 41, 47, 53, 59, 71, 83, 89, ... .

Your mission, should you decide to accept it, is to write a program to find the nth prime number in this arithmetic sequence for given positive integers a, d, and n.

Input

The input is a sequence of datasets. A dataset is a line containing three positive integers a, d, and n separated by a space. a and d are relatively prime. You may assume a <= 9307, d <= 346, and n <= 210.

The end of the input is indicated by a line containing three zeros separated by a space. It is not a dataset.

Output

The output should be composed of as many lines as the number of the input datasets. Each line should contain a single integer and should never contain extra characters.

The output integer corresponding to a dataset a, d, n should be the nth prime number among those contained in the arithmetic sequence beginning with a and increasing by d.

FYI, it is known that the result is always less than 106 (one million) under this input condition.

Sample Input

367 186 151

179 10 203

271 37 39

103 230 1

27 104 185

253 50 85

1 1 1

9075 337 210

0 0 0

Sample Output

92809

6709

12037

103

93523

14503

2

899429

      （1）编程思路。

      定义数组char flag[1000000]用于质数的判定。元素flag[i]==’1’表示整数i是质数；flag[i]==’0’表示整数i不是质数。

      对于输入的测试数据a、d和n，采用循环找到第n个质数在原等差数列中位置i。

       count = 0; 

       for(i=0;count<n;i++) 

          { 

           if(flag[a+d*i]=='1') 

              count++; 

          } 

      （2）源程序。

#include <iostream> 

using namespace std; 

int main() 

{ 

    int a,d,n,i,j,count; 

    char flag[1000000];

    for (i=2;i<1000000;i++)

              flag[i]='1';

       for(i=2;i<1000000;i++)      // 用筛法构建质数表

       {

        if (flag[i]=='1')

                     for (j=2*i;j<1000000;j+=i)

                            flag[j]='0';