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;
inline void in(int &n) {
  n = 0;
  int ch = getchar();
  int sign = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') sign = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    n = (n << 3) + (n << 1) + ch - '0', ch = getchar();
  }
  n = n * sign;
}
int main() {
  int n, t;
  scanf("%d %d", &n, &t);
  ;
  if (t <= 9) {
    for (int i = 1; i <= n; i++) {
      printf("%d", t);
    }
    printf("\n");
  } else {
    if (n == 1) {
      printf("-1\n");
    } else {
      printf("1");
      for (int i = 2; i <= n; i++) {
        printf("0");
      }
      printf("\n");
    }
  }
}
