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;
template <class Q>
Q _sqr(Q x) {
  return x * x;
}
template <class Q>
void gi(Q &x) {
  char ch = getchar();
  x = 0;
  int s = 1;
  while (ch < '0' || ch > '9') {
    if (ch == '-') s = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - 48, ch = getchar();
  if (s == -1) x = -x;
}
template <class Q>
void pi(Q x) {
  if (x < 0) {
    putchar('-');
    pi(-x);
  } else {
    if (x > 9) pi(x / 10);
    putchar(x % 10 + 48);
  }
}
const int sz = 100200;
const int huge = 1000000000;
const long long int hugell = 1ll << 61;
const double hugelf = 1e50;
const double eps = 1e-10;
const int ljz = 1000000007;
void MOD(int &x) {
  if (x >= ljz) x -= ljz;
  if (x < 0) x += ljz;
}
int Tm(int a, int b) {
  long long int x = a, y = b;
  return (x * y) % ljz;
}
int fpm(int x, int y) {
  int p = 1, s = x;
  for (; y; y >>= 1) {
    if (y & 1) p = Tm(p, s);
    s = Tm(s, s);
  }
  return p;
}
int Tm(int a, int b, int c) { return Tm(Tm(a, b), c); }
int Tm(int a, int b, int c, int d) { return Tm(Tm(a, b, c), d); }
int Tm(int a, int b, int c, int d, int e) { return Tm(Tm(a, b, c, d), e); }
int Tm(int a, int b, int c, int d, int e, int f) {
  return Tm(Tm(a, b, c, d, e), f);
}
int main() {
  int n, t;
  scanf("%d %d", &n, &t);
  if (t == 10) t = 1;
  if (t == 1 && n == 1) return printf("-1"), 0;
  printf("%d", t);
  n--;
  while (n--) putchar('0');
  return 0;
}