Codeforces Round #434 (Div. 2, based on Technocup 2018 Elimination Round 1)&&Codeforces 861A k-rounding【暴力】

A. k-rounding

time limit per test:1 second

memory limit per test:256 megabytes

input:standard input

output:standard output

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

题目链接：http://codeforces.com/contest/861/problem/A

题意： x的末尾有k个以上的0; 且x是n的倍数，x/(10^k) 是 n的倍数，x是10^k的倍数

下面给出AC代码：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll gcd(ll a,ll b)
{
    return b==0?a:gcd(b,a%b);
}
int main(void)
{
    ll n,k;
    cin>>n>>k;
    ll temp=1;
    for(ll i=1;i<=k;i++)
        temp*=10;
    cout<<n*temp/gcd(temp,n);
}