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;
string toString(long long num) {
  stringstream ss;
  ss << num;
  string str = ss.str();
  return str;
}
long long toLong(string str) {
  long long res;
  istringstream ss(str);
  ss >> res;
  return res;
}
int main() {
  long long n, t;
  cin >> n >> t;
  int num = 0;
  char s[105];
  for (int i = 0; i < n; i++) {
    s[i] = '0';
  }
  if (t == 10 && n == 1) {
    cout << "-1";
    return 0;
  } else {
    if (t == 10)
      s[0] = '1';
    else
      s[0] = t + '0';
  }
  s[n] = '\0';
  cout << s << endl;
  return 0;
}