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;
const long long MXN = 1e6 + 1;
const long long MNN = 1e3 + 1;
const long long MOD = 1e9 + 7;
const long long INF = 1e18;
const long long MAGIC = 1000;
const long double EPS = 1e-9;
vector<int> v;
int n, t;
int main() {
  cin >> n >> t;
  if (n == 1 && t == 10) {
    cout << -1;
    return 0;
  }
  if (t == 10) {
    cout << 1;
    for (int i = 1; i < n; i++) {
      cout << 0;
    }
    return 0;
  }
  string s;
  for (int i = 1; i <= 10; i++) {
    if (i % t == 0) {
      s += i + '0';
      if (i == 10) s += '0';
      break;
    }
  }
  while (n != 1) {
    s += "0";
    n--;
  }
  cout << s;
  return 0;
}