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
'''
[0, 1, 2, 3, 4]
[0, 1, 2, 3, 4]
[0, 1, 2, 3, 4]
[0, 0, 0]
[0, 0, 0]
[0, 0, 0]
[0, 0, 0]
[0, 0, 0]
'''
'''
a=[]
for i in range(3):
    t1=[]
    for j in range(5):
        t1.append(j)
    a.append(t1)
for i in a:
    print(i)
b=[[0 for i in range(3)] for j in range(5)]
for i in range(len(b)):
    for j in range(len(b[0])):
        b[i][j]=a[j][i]
for i in b:
    print(i)

res=[[0 for i in range(len(a))] for j in range(len(b[0]))]

for i in range(len(a)):
    for j in range(len(b[0])):
        for k in range(len(b)):
            res[i][j]+=(a[i][k]*b[k][j])
for i in res:
    print(i)
'''
n,t=map(int,input().strip().split())
minnum=int("1"+"0"*(n-1))
maxnum=int("1"+"0"*n)
ok=1
while(minnum<maxnum):
    if(minnum%t==0):
        ok=0
        break
    minnum+=1
if(not ok):
    print(minnum)
else:
    print(-1)
