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;
long long h, q, s, c, i, k, j, n, m, z, b, l, a, d, t, cas, i1;
bool myf2(int i, int j) { return (i > j); }
vector<vector<int> > vvk;
vector<int> va, vb, vk;
vector<pair<int, int> > vp;
string we1, we2, wex;
char zn;
int main() {
  cin >> n >> t;
  if (t == 10) {
    if (n == 1)
      cout << "-1" << endl;
    else {
      wex.assign(n, '1');
      wex[n - 1] = '0';
    }
  } else {
    zn = t + '0';
    wex.assign(n, zn);
  }
  cout << wex << endl;
  return 0;
}