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>
int main() {
  int n, i, t;
  char c;
  scanf("%d %d", &n, &t);
  char str[n];
  if (n == 1 && t != 10) {
    printf("%d", t);
    return 0;
  } else if (n == 1 && t == 10) {
    printf("-1");
    return 0;
  } else if (n >= 2 && t != 10) {
    c = t + '0';
    str[0] = c;
    for (i = 1; i < n; i++) {
      str[i] = '0';
    }
  } else {
    str[0] = '1';
    for (i = 1; i < n; i++) {
      str[i] = '0';
    }
  }
  str[n] = '\0';
  printf("%s", str);
  return 0;
}