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>
using namespace std;
int main() {
  int n, t;
  while (cin >> n >> t) {
    string s = "1";
    int i, l, x, c;
    l = s.length();
    if (n == 1 && t == 10) {
      cout << "-1" << endl;
      continue;
    } else if (n == 1) {
      cout << t << endl;
      continue;
    }
  a:
    c = 0;
    if (l < n) {
      for (i = l - 1; i >= 0; i--) {
        x = ((s[i] - 48) * t) + c;
        if (x >= 10) {
          c = x / 10;
          x %= 10;
        } else {
          c = 0;
        }
        s[i] = x + 48;
      }
      if (c == 1) {
        s.insert(0, "1");
      } else if (c == 2) {
        s.insert(0, "2");
      } else if (c == 3) {
        s.insert(0, "3");
      } else if (c == 4) {
        s.insert(0, "4");
      } else if (c == 5) {
        s.insert(0, "5");
      } else if (c == 6) {
        s.insert(0, "6");
      } else if (c == 7) {
        s.insert(0, "7");
      } else if (c == 8) {
        s.insert(0, "8");
      } else if (c == 9) {
        s.insert(0, "9");
      } else {
      }
      l = s.length();
      goto a;
    }
    cout << s << endl;
  }
}