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
#include <bits/stdc++.h>
using namespace std;
vector<long long> vec;
void primeFactors(long long n) {
  while (n % 2 == 0) {
    vec.push_back(2);
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      vec.push_back(i);
      n = n / i;
    }
  }
  if (n > 2) vec.push_back(n);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  long long d = 1;
  for (int i = 0; i < k; i++) {
    d *= 10;
  }
  primeFactors(n);
  vector<long long>::iterator it;
  vector<long long> vec2;
  int two = k, five = k;
  for (it = vec.begin(); it != vec.end(); it++) {
    if (*it == 5 && five != 0) {
      five--;
      continue;
    }
    if (*it == 2 && two != 0) {
      two--;
      continue;
    }
    vec2.push_back(*it);
  }
  long long result = 1;
  for (it = vec2.begin(); it != vec2.end(); it++) {
    result *= (*it);
  }
  cout << result * d;
}
