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 bPow(int v, int u, int mod) {
  if (!u) return 1;
  int res = bPow(v, u / 2, mod);
  res = (res * 1ll * res) % mod;
  if (u & 1) res = (res * 1ll * v) % mod;
  return res;
}
bool fb(long long x) { return 0; }
int bSearch(int l, int r) {
  while (r - l > 1) {
    int m = (l + r) / 2;
    if (fb(m))
      l = m;
    else
      r = m;
  }
  if (fb(l + 1)) l++;
  if (!fb(l)) l = 0;
  return l;
}
void faster() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  if (0) {
    freopen(
        "input"
        ".in",
        "r", stdin);
    freopen(
        "input"
        ".out",
        "w", stdout);
  }
}
vector<int> prime;
bool notprime[500010];
void resheto() {
  int n = 500010 - 10;
  for (int i = 2; i < n; i++)
    if (!notprime[i]) {
      prime.push_back(i);
      for (int j = i; j < n; j += i) notprime[j] = 1;
    }
}
int is_comp(long long p) {
  for (int i = 0; prime[i] * 1ll * prime[i] <= p; i++) {
    if (p % prime[i] == 0) return prime[i];
  }
  return 0;
}
int is_prime(long long p) {
  for (int i = 0; prime[i] * 1ll * prime[i] <= p; i++) {
    if (p % prime[i] == 0) return 0;
  }
  return 1;
}
int gcd(int a, int b) {
  if (a > b) swap(a, b);
  if (!a) return b;
  return gcd(b % a, a);
}
int n, k;
int main() {
  faster();
  cin >> n >> k;
  k = bPow(10, k, 1000000007);
  cout << 1ll * n * k / gcd(n, k);
  return 0;
}
