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 int N = 100100;
const double pi = 3.141592653589793238462643383279502884197169399375105;
set<int> st;
vector<pair<int, int>> vp;
map<char, long long> mpx;
map<long long, string> mpy;
int farr[N];
vector<int> v;
vector<int> v1;
vector<int> v2;
vector<int> v3;
priority_queue<int> pq;
int main() {
  long long n, t;
  string str = "";
  cin >> n >> t;
  if (t == 10) {
    if (n == 1) {
      cout << -1;
    } else {
      n--;
      while (n--) {
        str += '1';
      }
      str += '0';
      cout << str << endl;
    }
  } else {
    string tt = to_string(t);
    while (n--) str += tt;
    cout << str << endl;
  }
  return 0;
}
