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 itertools # to handle the long shit in c 
#from timeit import default_timer

info = raw_input().split(' ')
n = int(info[0])
t = int(info[1])

#print 'n : ', n
#print 't : ', t

#start_time = default_timer()

first_number = ''
for i in xrange(n):
	if i+1 == 1:
		first_number += str('1')
	else:
		first_number += str('0')
first_number = int(first_number)

last_number = ''
for j in xrange(n):
	last_number += str('9')
last_number = int(last_number)

#print 'first number : ', first_number
#print 'last number : ', last_number

foundIt = False
for number in itertools.count(first_number,1):
	#print 'number : ', number
	if len(str(number)) > n :
		break
	if t > last_number:
		break
	if number % t == 0:
		print number
		foundIt = True
		#end_time = default_timer()
		#print 'time spent : ', end_time - start_time
		break

if foundIt == False:
	#end_time = default_timer()
	print -1
	#print 'time spent : ', end_time - start_time
