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>
#pragma GCC optimize("Ofast,fast-math")
std::mt19937_64 rng(
    std::chrono::high_resolution_clock::now().time_since_epoch().count());
inline void setup() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
}
inline void solve();
int main(int argc, char* argv[]) {
  setup();
  long long t = 1;
  while (t--) solve();
  return 0;
}
using namespace std;
inline void solve() {
  long long n, t;
  cin >> n >> t;
  if (t == 10) {
    if (n == 1)
      cout << -1;
    else {
      for (long long i = 0; i < n - 1; ++i) cout << 1;
      cout << 0;
    }
  } else {
    for (long long i = 0; i < n; ++i) cout << t;
  }
}