A - k-rounding

A - 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.

Examples

Input

375 4

 

Output

30000

 

Input

10000 1

 

Output

10000

 

Input

38101 0

 

Output

38101

 

Input

123456789 8

 

Output

12345678900000000

 

题目描述：

对一个给定的数n，进行k-round操作就是，找出一个最小的数，这个数可以被n整除，而且要保证这个数后面至少有k个0。

分析：

就是求n和10^k的最小公倍数。使用公式a*b/gcd(a,b)。因为这个数会很大，而且他一定要乘以10^k。所以先算出n/gcd(n,10^k)，再在后面打印k个0即可。

代码：

#include<iostream> 
#include<algorithm>
#include<cmath>
#define MAX(x,y) x>y?x:y;
using namespace std;
typedef long long ll;
int gcd(int a,int b)
{
    if(b==0)
    {
        return a;
    }
    gcd(b,a%b);
}
int main()
{
    int n,k;
    scanf("%d %d",&n,&k);
    int p=pow(10,k);
    ll lcm=n/gcd(n,p);
    printf("%lld",lcm);
    for(int i=0;i<k;i++) putchar('0');
    putchar('\n');
    return 0;
}