題干：

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.

Examples

Input

5 4 5

Output

524848

Input

12 11 1

Output

121

Input

260 150 10

Output

-1

解題報(bào)告：

? ? 先判斷第一個(gè)可以不可以，如果不可以輸出-1，如果可以就后面都輸出0。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define ll long long #define pb push_back #define pm make_pair #define fi first #define se second using namespace std; const int MAX = 2e5 + 5; ll a,b,n; ll cur; int main() {cin>>a>>b>>n;int flag = 0;for(int i = 0; i<=9; i++) {cur = a*10+i;if(cur%b == 0) {flag = 1;break;}}if(flag == 0) {puts("-1");return 0 ;}n--;printf("%lld",cur);for(int i = 1; i<=n; i++) putchar('0');return 0 ;}

總結(jié)：

? wjh大佬寫(xiě)了dfs找到循環(huán)節(jié)然后再輸出的。

總結(jié)

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