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 a[11][105];
int main() {
  int n, t;
  for (int i = 2; i <= 10; i++) {
    a[i][1] = 1;
    for (int j = 2; j <= 100; j++) a[i][j] = (a[i][j - 1] * 10 + 1) % i;
  }
  scanf("%d%d", &n, &t);
  if (n == 1 && t == 10)
    puts("-1");
  else if (t == 10) {
    for (int i = 0; i < n - 1; i++) printf("1");
    puts("0");
  } else {
    int x = a[t][n - 1] * 10;
    x %= t;
    for (int i = 0; i < n - 1; i++) printf("1");
    printf("%d\n", t - x);
  }
  return 0;
}