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
#include <bits/stdc++.h>
using namespace std;
const double eps = 1e-9;
const long long MOD = 998244353;
const long long N = (long long)1e5 + 9;
const long long inf = 1 << 30;
template <class T>
T gcd(T a, T b) {
  return (b != 0 ? gcd<T>(b, a % b) : a);
}
template <class T>
T lcm(T a, T b) {
  return (a / gcd<T>(a, b) * b);
}
template <typename T>
T exp(T b, T p) {
  T x = 1;
  while (p) {
    if (p & 1) x = (x * b);
    b = (b * b);
    p = p >> 1;
  }
  return x;
}
inline long long ones(long long n) {
  long long res = 0;
  while (n && ++res) n -= n & (-n);
  return res;
}
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(0);
  long long n, t;
  cin >> n >> t;
  if (n == 1 and t == 10) return cout << -1, 0;
  if (t == 10) return cout << 1 << string((n - 1), '0'), 0;
  cout << string(n, t + '0');
}
