Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.

Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print  - 1.

Input

The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.

Output

Print one such positive number without leading zeroes, — the answer to the problem, or  - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.

Examples

Input

3 2


Output

712
#include <bits/stdc++.h>
using ull = unsigned long long;
using ll = long long;
using namespace std;
ll dmod = 1e9 + 7;
ll fpow(ll a, ll b, ll mod) {
  ll x = 1;
  while (b != 0) {
    if (b & 1) {
      x *= a;
      x %= mod;
    }
    a = a * a;
    a %= mod;
    b >>= 1;
  }
  return x;
}
void solve() {
  ll n, t;
  cin >> n >> t;
  if (t == 10 && n == 1) {
    cout << -1 << "\n";
    return;
  }
  if (t == 10) {
    n--;
  }
  cout << t;
  for (ll i = 0; i < n - 1; ++i) {
    cout << 0;
  }
  cout << "\n";
}
void start() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.precision(10);
}
int main() {
  start();
  ll t = 1;
  while (t--) {
    solve();
  }
}
