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;
void debug(string j) { cout << "| Value : " << j << " |\n"; }
void debug(int j) { cout << "| Value : " << j << " |\n"; }
void debug(int i, string j) { cout << "Iteration " << i << ": " << j << "\n"; }
void debug(int i, int j) { cout << "Iteration " << i << ": " << j << "\n"; }
void testcase() {
  int n, k;
  cin >> n >> k;
  if (k == 10 && n < 2)
    printf("-1");
  else if (k == 10) {
    cout << 1;
    for (int i = 0; i < n - 1; i++) cout << 0;
  } else
    for (int i = 0; i < n; i++) cout << k;
  puts("");
}
int main() {
  int t = 1;
  for (int i = 1; i <= t; i++) {
    testcase();
  }
  return 0;
}