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>
long long int MAX(long long int a, long long int b) {
  if (a > b)
    return a;
  else
    return b;
}
using namespace std;
long long int MIN(long long int a, long long int b) {
  if (a < b)
    return a;
  else
    return b;
}
long long int MOD(long long int a, long long int b) {
  return a >= 0 ? a % b : (b - abs(a % b)) % b;
}
int X[] = {1, 0, -1, 0, 1, -1, -1, 1};
int Y[] = {0, 1, 0, -1, 1, 1, -1, -1};
int main() {
  int n, t;
  scanf("%d%d", &n, &t);
  if (n == 1 && t == 10)
    printf("-1\n");
  else {
    string str;
    if (t < 10)
      n--;
    else
      n -= 2;
    printf("%d", t);
    for (int x = 0; x < n; x++) {
      str += ('0');
    }
    cout << str;
    printf("\n");
  }
  return 0;
}