codeforces 858 A k-rounding(数学)

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 10and 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

计算出n*x= y* 1ek

结果一定包含k个0 ，一定可以取余n 那么答案最小值就是n*1ek/gcd(n，1ek)；

从宏观角度 列出表达式 观察计算结果

#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <algorithm>#include <climits>using namespace std;const int N = 1000000;typedef long long LL;LL d[20];vector<LL>p;int main(){    d[0]=1;    for(int i=1; i<=18; i++) d[i]=d[i-1]*10;    LL n, k;    scanf("%I64d %I64d", &n, &k);    printf("%I64d\n",n*d[k]/__gcd(n,d[k]));    return 0;}