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, t, z, q, m, c, k;
  int a[1000];
  scanf("%d %d", &n, &t);
  if (n == 1 && t == 10) {
    printf("-1");
  } else if (t == 10) {
    z = n - 1;
    for (c = 0; c < z; c++) {
      a[c] = 1;
    }
    a[z] = 0;
    for (c = 0; c < n; c++) {
      printf("%d", a[c]);
    }
  } else {
    for (c = 0; c < n; c++) {
      a[c] = t;
    }
    for (c = 0; c < n; c++) {
      printf("%d", a[c]);
    }
  }
  return 0;
}