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;
const int inf = 1 << 28;
int n, a, b, c;
int dp[200000];
int solve(int total, int count, int parent = 0) {
  if (total < 0) return -inf;
  int &ret = dp[total];
  if (ret != -1) return ret;
  if (total == 0) return count;
  int aa = solve(total - a, count + 1, a);
  int bb = solve(total - b, count + 1, b);
  int cc = solve(total - c, count + 1, c);
  ret = max(aa, max(bb, cc));
  return ret;
}
int gcd(int a, int n) {
  if (a == n || n == 0 || a == 0)
    return a;
  else if (a > n)
    return gcd(a - n, n);
  else
    return gcd(a, n - a);
}
int level(int i) {
  int x = 0;
  for (int j = 1; j <= i; j++) {
    x += j;
  }
  return x;
}
int main() {
  int n, t;
  cin >> n >> t;
  if (t < 10) {
    for (int i = 0; i < n; i++) {
      cout << t;
    }
  } else {
    if (n < 2) {
      cout << -1;
      return 0;
    }
    cout << 1;
    for (int i = 0; i < n - 1; i++) {
      cout << 0;
    }
  }
  return 0;
}