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, i, k, l, flag = 0;
  scanf("%d %d", &n, &t);
  for (i = 0; i < n; i++) {
    if (t == 2) {
      printf("%d", 2);
    }
    if (t == 3) {
      printf("%d", 3);
    }
    if (t == 4) {
      printf("%d", 4);
    }
    if (t == 5) {
      printf("%d", 5);
    }
    if (t == 6) {
      printf("%d", 6);
    }
    if (t == 7) {
      printf("%d", 7);
    }
    if (t == 8) {
      printf("%d", 8);
    }
    if (t == 9) {
      printf("%d", 9);
    }
  }
  if (t != 10) {
    printf("\n");
  }
  if (n == 1 && t == 10) {
    printf("-1\n");
  }
  if (n > 1 && t == 10) {
    printf("%d", 1);
    for (i = 1; i < n; i++) {
      printf("%d", 0);
    }
    printf("\n");
  }
  return 0;
}