CodeForces

k-rounding

For a given positive integer n denote its k-rounding as the minimum positive integer x, such that x ends with k or more zeros in base 10 and is divisible by n.


For example, 4-rounding of 375 is 375·80 = 30000. 30000 is the minimum integer such that it ends with 4 or more zeros and is divisible by 375.


Write a program that will perform the k-rounding of n.


Input
The only line contains two integers n and k (1 ≤ n ≤ 109, 0 ≤ k ≤ 8).


Output
Print the k-rounding of n.


Example
Input
375 4
Output
30000
Input
10000 1
Output
10000
Input
38101 0
Output
38101
Input
123456789 8
Output
12345678900000000

题意：对于一个数字n,定义它的"k阶数"为最小的正整数x,这里x以k个或者更多的0结尾,且能够被n整除. 比如375的4阶数为 375*80 = 30000.

#include<stdio.h>#include<iostream>#include<algorithm>#include<math.h>using namespace std;int main(){    long long n,k,m=1;    cin>>n>>k;    while(k--)        m*=10;    cout<<(n*m)/__gcd(n,m)<<endl;}