A. Adding Digits

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Vasya has got two number: a and b. However, Vasya finds number a too short. So he decided to repeat the operation of lengthening number a n times.

One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number b. If it is impossible to obtain the number which is divisible by b, then the lengthening operation cannot be performed.

Your task is to help Vasya and print the number he can get after applying the lengthening operation to number a n times.

Input

The first line contains three integers: a, b, n (1 ≤ a, b, n ≤ 10⁵).

Output

In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number a n times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.

Sample test(s)

input

5 4 5

output

524848

input

12 11 1

output

121

input

260 150 10

output

-1

解题说明：此题的意思是在a后面添加n个数字，确保添加的每一个数字后组成的新数都能被b整除。这里可以取个巧，因为任何被b整除的数字后面添加0以后肯定能继续被b整除，所以我们只需要找到第一个添加的数字该是多少即可，一个循环加以判断，后面添上n-1个0.如果第一个数字找不到，那就输出-1

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;

int main() 
{
	int a,b,n,i,j,flag;
	flag=0;
	scanf("%d %d %d",&a,&b,&n);
	for(i=0;i<10;i++)
	{
		if((a*10+i)%b==0)
		{
			flag=1;
			printf("%d",a);
			printf("%d",i);
			for(j=0;j<n-1;j++)
			{
				printf("0");
			}
			printf("\n");
			break;
		}
	}
	if(flag==0)
	{
		printf("-1\n");
	}
	return 0;
}