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
#include <bits/stdc++.h>
using namespace std;
int n, m, i, j, k, l, a[1000009], b[1000009], p[1000009], ans, cn, t, x, y, z,
    mx, mn, s;
char c[1000009], d[1000009], ch;
int main() {
  while (scanf("%d %d", &n, &t) != EOF) {
    if ((n == 1) && (t == 10)) {
      printf("-1\n");
    } else if (t < 10) {
      for (i = 0; i < n; i++) {
        printf("%d", t);
      }
      printf("\n");
    } else {
      for (i = 0; i < n - 1; i++) {
        printf("%d", t - 1);
      }
      printf("0\n");
    }
  }
  return 0;
}
