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>
#pragma GCC optimize("O2")
using namespace std;
const char nl = '\n';
const long long inf = 1e9;
const long long inf2 = 1e18 + 10;
const long double inf3 = 1e14;
const long long mod = 998244353;
const long double eps = 1e-6;
const int logn = 18, maxn = 1e5, maxm = 1e5 + 5;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n;
  cin >> n;
  int t;
  cin >> t;
  if (n == 1 && t == 10) {
    cout << -1;
    return 0;
  }
  cout << t;
  for (int i = 0; i < (n - 1 - t / 10); i++) cout << 0;
}
