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;
map<char, int> mp;
map<char, int> mp1;
map<char, int> mp2;
long long frq[100002] = {};
bool lol = true;
int main() {
  bool u = false;
  long long x, y, p;
  scanf("%lld%lld", &x, &y);
  long long num = y, sum = 0;
  if (x == 1 && y == 10) {
    printf("-1\n");
    return 0;
  }
  if (y < 10) {
    for (int i = 1; i <= x; i++) {
      cout << num;
    }
  } else {
    for (int i = 1; i <= x; i++) {
      if (i == x) {
        cout << num % 10;
      } else
        cout << num / 10;
    }
  }
  return 0;
}
