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;
  scanf("%d%d", &n, &t);
  int i;
  long long s = 1;
  if (n == 1) {
    if (t == 10) {
      printf("-1");
    } else if (t == 1 || t == 2 || t == 3 || t == 4 || t == 5 || t == 6 ||
               t == 7 || t == 8 || t == 9) {
      printf("%d", t);
    }
  }
  if (n != 1) {
    if (t % 6 == 0) {
      for (i = 1; i <= n; i++) {
        printf("6");
      }
    } else if (t % 2 == 0 && t / 2 != 5) {
      for (i = 1; i <= n; i++) {
        printf("8");
      }
    } else if (t % 10 == 0) {
      for (i = 2; i <= n; i++) {
        printf("1");
      }
      printf("0");
    } else if (t % 9 == 0) {
      for (i = 1; i <= n; i++) {
        printf("9");
      }
    } else if (t % 3 == 0) {
      for (i = 1; i <= n; i++) {
        printf("3");
      }
    } else if (t % 5 == 0) {
      for (i = 1; i <= n; i++) {
        printf("5");
      }
    } else if (t % 6 == 0) {
      for (i = 0; i <= n; i++) {
        printf("6");
      }
    } else if (t % 7 == 0) {
      for (i = 1; i <= n; i++) {
        printf("7");
      }
    }
  }
}