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>
#pragma comment(linker, "/STACK:1677721600")
using namespace std;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double ee = exp(1.0);
const int inf = 0x3f3f3f3f;
const int maxn = 1e6 + 10;
const double pi = acos(-1.0);
const long long iinf = 0x3f3f3f3f3f3f3f3f;
int readT() {
  char c;
  int ret = 0, flg = 0;
  while (c = getchar(), (c < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    flg = 1;
  else
    ret = c ^ 48;
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c ^ 48);
  return flg ? -ret : ret;
}
long long readTL() {
  char c;
  int flg = 0;
  long long ret = 0;
  while (c = getchar(), (c < '0' || c > '9') && c != '-')
    ;
  if (c == '-')
    flg = 1;
  else
    ret = c ^ 48;
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c ^ 48);
  return flg ? -ret : ret;
}
int main() {
  int n = readT();
  int t = readT();
  if (t == 10) {
    if (n >= 2) {
      printf("1");
      for (int i = (1); i <= (n - 1); i++) {
        printf("0");
      }
    } else {
      printf("-1");
    }
    return 0;
  }
  for (int i = (1); i <= (n); i++) {
    printf("%d", t);
  }
  return 0;
}
