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;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
string toBin(long long a) {
  string res = "";
  while (a) {
    res += char((a & 1) + '0');
    a >>= 1;
  }
  reverse(res.begin(), res.end());
  return res;
}
const int mod = 1e9 + 7;
long long expo(long long base, long long pow) {
  long long ans = 1;
  while (pow != 0) {
    if (pow & 1 == 1) {
      ans = ans * base;
      ans = ans % mod;
    }
    base *= base;
    base %= mod;
    pow /= 2;
  }
  return ans;
}
double pi = 3.141592653589793238462643;
double error = 0.0000001;
const int M = 100001;
int main() {
  int n, t;
  scanf("%d", &(n));
  scanf("%d", &(t));
  if (n == 1 && t == 10) {
    cout << "-1";
    return 0;
  }
  if (t < 10) {
    for (int i = 0; i < n; i++) {
      cout << t;
    }
    return 0;
  }
  for (int i = 0; i < n - 1; i++) {
    cout << "1";
  }
  cout << "0";
}