Problem: 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

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <typename... Args>
void logger(string vars, Args &&...values) {
  cout << vars << " = ";
  string delim = "";
  (..., (cout << delim << values, delim = ", "));
  cout << endl;
}
vector<long long> invi(long long n) {
  long long x;
  vector<long long> a;
  for (long long i = 0; i < n; i++) {
    cin >> x;
    a.push_back(x);
  }
  return a;
}
vector<bool> prime;
void checkPrime(long long n) {
  for (long long i = 0; i < n + 1; i++) {
    prime.push_back(true);
  }
  prime[0] = false;
  prime[1] = false;
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  while (t--) {
    solve();
  }
}
void solve() {
  long long n, t;
  cin >> n >> t;
  if (t == 10) {
    if (n == 1) {
      cout << "-1"
           << "\n";
      return;
    }
    n--;
    while (n) {
      cout << 1;
      n--;
    }
    cout << 0 << "\n";
  } else {
    while (n) {
      cout << t;
      n--;
    }
  }
}