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 cmpfunc(const void *a, const void *b) { return (*(int *)a - *(int *)b); }
long long int gcd(long long int m, long long int n) {
  if (n == 0) return m;
  return gcd(n, m % n);
}
long long int fastpow(long long int a, long long int b, long long int m) {
  long long int r = 1;
  while (b > 0) {
    if (b % 2 == 1) r = (r * a) % m;
    b = b >> 1;
    a = (a * a) % m;
  }
  return r;
}
int prime(long long int x) {
  if (x == 1) return 0;
  if (x <= 3) return 1;
  if (x % 6 == 1 || x % 6 == 5) {
    long long int y = sqrt(x), i;
    for (i = 2; i <= y; i++)
      if (x % i == 0) return 0;
    return 1;
  }
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long int n, t;
  cin >> n >> t;
  if (n == 1 && t == 10)
    cout << -1 << endl;
  else if (t <= 9) {
    for (long long int i = 0; i < n; i++) cout << t;
    cout << endl;
  } else {
    cout << 1;
    for (long long int i = 1; i < n; i++) cout << 0;
    cout << endl;
  }
  return 0;
}