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;
int countDigit(long long n) {
  int count = 0;
  while (n != 0) {
    n = n / 10;
    ++count;
  }
  return count;
}
int main() {
  long long c = 1;
  long long n, k;
  cin >> n >> k;
  bool test = true;
  int count = 0;
  int t = countDigit(k);
  vector<char> s;
  if (n < t) {
    cout << -1 << endl;
    return 0;
  }
  if (k != 10) {
    for (int i = 0; i < n; i++) {
      s.push_back(k + '0');
    }
  } else {
    s.push_back('1');
    for (int i = 0; i < n - 1; i++) {
      s.push_back('0');
    }
  }
  for (int i = 0; i < s.size(); i++) {
    cout << s[i];
  }
  cout << endl;
}