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 n, t;
int main() {
  cin >> n >> t;
  if (n == 1 && t == 10) {
    cout << -1;
    return 0;
  }
  if (t == 2) {
    for (int i = 0; i < n - 1; i++) {
      cout << 9;
    }
    cout << 2;
    return 0;
  }
  if (t == 3) {
    for (int i = 0; i < n - 1; i++) {
      cout << 9;
    }
    cout << 3;
    return 0;
  }
  if (t == 4) {
    for (int i = 0; i < n - 1; i++) {
      cout << 4;
    }
    cout << 4;
    return 0;
  }
  if (t == 5) {
    for (int i = 0; i < n - 1; i++) {
      cout << 9;
    }
    cout << 5;
    return 0;
  }
  if (t == 6) {
    for (int i = 0; i < n - 1; i++) {
      cout << 9;
    }
    cout << 6;
    return 0;
  }
  if (t == 7) {
    for (int i = 0; i < n - 1; i++) {
      cout << 7;
    }
    cout << 7;
    return 0;
  }
  if (t == 8) {
    for (int i = 0; i < n - 1; i++) {
      cout << 8;
    }
    cout << 8;
    return 0;
  }
  if (t == 9) {
    for (int i = 0; i < n - 1; i++) {
      cout << 9;
    }
    cout << 9;
    return 0;
  }
  if (t == 10) {
    for (int i = 0; i < n - 1; i++) {
      cout << 9;
    }
    cout << 0;
    return 0;
  }
  return 0;
}