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;
const double pi = 2 * acos(0.0);
const int INF = 0x3f3f3f3f;
const double inf = 1.0 / 0.0;
void solve();
void Adi() {
  ios::sync_with_stdio(false);
  cin.tie(0);
}
int main() {
  Adi();
  solve();
  cerr << "time taken : " << (float)clock() / CLOCKS_PER_SEC << " secs"
       << "\n";
  return 0;
}
void solve() {
  long long n, k, i, a, flag = 0;
  cin >> n >> k;
  if (k < 10) {
    while (n--) cout << k;
  } else {
    if (k == 10 && n == 1)
      cout << "-1";
    else {
      cout << 1;
      n--;
      while (n--) cout << 0;
    }
  }
}
