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.

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.

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.

#include 
    
     #include 
     
      #include 
      
       #define N 1000000int prime[N];void init(){    int i, j, k;    for (i = 0; i < N; i++){        prime[i] = 1;    }    prime[0] = 0;    prime[1] = 0;    for (i = 0; i < N; i++){        if (prime[i] != 0){            for (j = i + i; j < N; j += i){                prime[j] = 0;            }        }    }}int main(void){    int a, d, n;    int times, i, tmp;    init();    while (scanf("%d%d%d", &a, &d, &n), a != 0 || d != 0 || n != 0){        times = 0;        for (i = a; i < 1000000 && times != n; i += d){            if (prime[i] != 0){                times++;                tmp = i;            }        }        printf("%d\n", tmp);    }    return 0;}