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 N = 2e5 + 5;
const int mod = 1e8;
long long a[N], b[N], c[N], d, dp[N], dis;
char o;
pair<int, int> p;
bool mark[N], h = false;
string s;
vector<int> vt;
deque<int> dq;
int main() {
  long long t, n, m, f, sum = 0, k, ans, c1, c2, w1, w2;
  cin >> n >> t;
  if (n == 1 && t == 10) {
    cout << -1 << endl;
  } else if (t == 10) {
    cout << t;
    for (int i = 0; i < n - 2; i++) {
      cout << 0;
    }
  } else {
    cout << t;
    for (int i = 0; i < n - 1; i++) {
      cout << 0;
    }
  }
}