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;
using ll = long long;
ll md = 1000000007;
ll exp(ll a, ll b) {
  ll r = 1ll;
  while (b > 0) {
    if (b & 1) {
      r = r * (a % md);
      r = (r + md) % md;
    }
    b >>= 1;
    a = (a % md) * (a % md);
    a = (a + md) % md;
  }
  return (r + md) % md;
}
ll gcd(ll a, ll b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
ll Min(ll a, ll b) {
  if (a < b) return a;
  return b;
}
ll Max(ll a, ll b) {
  if (a > b) return a;
  return b;
}
const double PI = 3.1415926535897932238460;
ll const Mxn = 1e9;
int main() {
  long long n;
  int k;
  cin >> n >> k;
  long long a = n;
  int cnt5 = 0;
  while (a % 5 == 0) a /= 5, cnt5++;
  a = n;
  int cnt2 = 0;
  while (a % 2 == 0) a /= 2, cnt2++;
  while (cnt5 < k) n *= 5, cnt5++;
  while (cnt2 < k) n *= 2, cnt2++;
  cout << n << "\n";
}
