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;
template <class A, class B>
ostream& operator<<(ostream& o, const pair<A, B>& p) {
  return o << "(" << p.first << ", " << p.second << ")";
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "[";
  for (__typeof((v).end()) it = (v).begin(); it != (v).end(); ++it)
    o << *it << ", ";
  return o << "]";
}
template <class T>
ostream& operator<<(ostream& o, const set<T>& v) {
  o << "[";
  for (__typeof((v).end()) it = (v).begin(); it != (v).end(); ++it)
    o << *it << ", ";
  return o << "]";
}
template <class T>
inline void MAX(T& a, T b) {
  if (a < b) a = b;
}
template <class T>
inline void MIN(T& a, T b) {
  if (a > b) a = b;
}
template <class T>
inline bool read(T& x) {
  int c = getchar();
  int sgn = 1;
  while (~c && c < '0' || c > '9') {
    if (c == '-') sgn = -1;
    c = getchar();
  }
  for (x = 0; ~c && '0' <= c && c <= '9'; c = getchar()) x = x * 10 + c - '0';
  x *= sgn;
  return ~c;
}
int main() {
  int n, t;
  cin >> n >> t;
  int k = 0;
  int m = t;
  while (m > 0) {
    m /= 10;
    k++;
  }
  if (m > 0) {
    k++;
  }
  if (k > n) {
    cout << -1;
    return 0;
  }
  cout << t;
  while (k < n) {
    cout << 0;
    k++;
  }
}
