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;
int main() {
  int n, t;
  scanf("%d%d", &n, &t);
  if (n == 1) {
    if (t < 10)
      printf("%d\n", t);
    else
      printf("-1\n");
  } else {
    if (t == 10) {
      printf("1");
      for (int i = 1; i <= (n - 1); i++) printf("0");
      printf("\n");
    } else {
      int r = 1;
      for (int i = 1; i <= (n - 1); i++) r = r * 10 % t;
      printf("1");
      for (int i = 1; i <= (n - 2); i++) printf("0");
      printf("%d\n", t - r);
    }
  }
  return 0;
}