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;
int nod1(int a, int b) { return a ? nod1(b % a, a) : b; }
int nod(int a, int b) {
  if (a > b) swap(a, b);
  return nod1(a, b);
}
int gcd(int a, int b) {
  if (a > b) swap(a, b);
  return nod1(a, b);
}
int nok(int a, int b) { return a / nod(a, b) * b; }
long long sum(vector<long long> v) {
  long long ans = 0;
  for (int i = 0; i < v.size(); i++) ans += v[i];
  return ans;
}
long long sum(vector<int> v) {
  long long ans = 0;
  for (int i = 0; i < v.size(); i++) ans += v[i];
  return ans;
}
bool letter(char c) { return 'a' <= c && c <= 'z'; }
bool LETTER(char c) { return 'A' <= c && c <= 'Z'; }
string intToStr1(long long n) {
  return n ? intToStr1(n / 10) + (char)('0' + n % 10) : "";
}
string intToStr(long long n) { return n ? intToStr(n) : "0"; }
long long strToInt(string s) {
  long long ans = 0;
  for (int i = 0; i < s.size(); i++) ans = 10 * ans + (s[i] - '0');
  return ans;
}
bool digit(char c) { return '0' <= c && c <= '9'; }
bool prime(int n) {
  if (n == 1) return 0;
  for (int i = 2; i * i <= n; i++)
    if (n % i == 0) return 0;
  return 1;
}
const double zero = 0;
const long long null = 0;
const long long INF = 1e18;
const long long MOD = 1000000007;
const double PI = atan2(0, -1);
const int MAXN = 200001;
const double eps = 1e-8;
const int dx[4] = {-1, 0, 1, 0};
const int dy[4] = {0, 1, 0, -1};
vector<long long> a;
int main() {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  for (int i = 0; i < k; i++) {
    if (n % 2 == 0) n /= 2;
    if (n % 5 == 0) n /= 5;
  }
  cout << n;
  for (int i = 0; i < k; i++) cout << 0;
  return 0;
}
