/*
    Find the prime factors of positive integers.
    The numbers are given on the command line or standard input (one to a line).
    The output is of the form  
    
        number=factor*factor*factor.
    
    Naive algorithm.  (It's fast enough.)
*/

#include <errno.h>
#include <limits.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

unsigned long smallest_divisor(unsigned long n)
{
    unsigned long i;
    unsigned long limit = (unsigned long) sqrt(n);
    
    if (n <= 2)
        return n;
    if (2 <= limit && n % 2 == 0)
        return 2;
    if (3 <= limit && n % 3 == 0)
        return 3;
    for (i = 5; i <= limit; i += 6) {
        if (n % i == 0)
            return i;
        if (n % (i + 2) == 0)
            return i + 2;
    }
    return n;
}

static void factor(const char *number)
{
    char *end;
    unsigned long n = (errno = 0, strtoul(number, &end, 10));
    if (n == ULONG_MAX 
	|| (n == 0 && errno != 0)
	|| end[strspn(end, " \t\n")] != '\0') {
        fprintf(stderr, "%s: not an unsigned long\n", number);
        exit(1);
    }
    printf("%lu=", n);
    for (;;) {
        unsigned long divisor = smallest_divisor(n);
        printf("%lu", divisor);
        if (divisor == n) 
            break;
        printf("*");
        n /= divisor;
    }
    printf("\n");
}

int main(int argc, char **argv)
{
    if (argc == 1) {
        char line[81];
        while (fgets(line, sizeof line, stdin))
            factor(line);
    } else {
        int i;
        for (i = 1; i < argc; ++i)
            factor(argv[i]);
    }
    return 0;
}