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 double PI =
    3.14159265358979323846264338327950288419716939937510582097494459230;
const long long MOD = 1000000007LL;
const long long INF = 1e9;
const long long LINF = 1e18;
const long double EPS = 1e-9;
const long double GOLD = ((1 + sqrt(5)) / 2);
template <class T>
int getbit(T s, int i) {
  return (s >> 1) & 1;
}
template <class T>
T onbit(T s, int i) {
  return s | (T(1) << i);
}
template <class T>
T offbit(T s, int i) {
  return s & (~(T(1) << i));
}
template <class T>
int cntbit(T s) {
  return __builtin_popcount(s);
}
long long n, t;
void VarInput() { cin >> n >> t; }
void ProSolve() {
  if (n == 1 && t == 10)
    cout << "-1";
  else {
    if (t == 10)
      cout << "1";
    else
      cout << t;
    for (long long i = 1; i < n; i++) cout << "0";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  VarInput();
  ProSolve();
  return 0;
}