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
import sys
from random import choice,randint
inp=sys.stdin.readline
out=sys.stdout.write
flsh=sys.stdout.flush
 
sys.setrecursionlimit(10**9)
inf = 10**20
eps = 1.0 / 10**10
mod = 10**9+7
dd = [(-1,0),(0,1),(1,0),(0,-1)]
ddn = [(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]
 
def MI(): return map(int, inp().strip().split())
def LI(): return list(map(int, inp().strip().split()))
def LLI(): return [list(map(int, l.split())) for l in sys.stdin.readlines().strip()]
def LI_(): return [int(x)-1 for x in inp().strip().split()]
def LF(): return [float(x) for x in inp().strip().split()]
def LS(): return inp().strip().split()
def I(): return int(inp().strip())
def F(): return float(inp().strip())
def S(): return inp().strip()
def pf(s): return out(s+'\n')
def JA(a, sep): return sep.join(map(str, a))
def JAA(a, s, t): return s.join(t.join(map(str, b)) for b in a)

def main():
    n,t=MI()
    if t==10 and n==1:
        pf("-1")
    elif t<10:
        pf(str(t)*n)
    else:
        pf("1"+"0"*(n-1))

if __name__ == "__main__":
    main()