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 binSearch(long long s[], long long l, long long r, long long t) {
  long long i = l;
  while (l <= r) {
    long long m = (l + r + 1) / 2;
    if (s[m] - s[i - 1] <= t && s[m + 1] - s[i - 1] > t)
      return m;
    else if (s[m] - s[i - 1] < t)
      l = m + 1;
    else
      r = m - 1;
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long i, n, k, j, t;
  cin >> n >> t;
  if (t != 10) {
    cout << t;
    for (i = 0; i < n - 1; i++) {
      cout << 0;
    }
  } else {
    if (n == 1) {
      cout << -1;
      return 0;
    }
    cout << 10;
    for (i = 0; i < n - 2; i++) {
      cout << 0;
    }
  }
  return 0;
}