'''
K+1個のブロック
a_0,...,a_K
a_i>=1

a_{2i}個の0
a_{2i+1}個の1
辞書順最小にするためには
a_0を極力大きく
a_2,a_4,...,a_{2i}を1
'''
def solve(N,M,K):
    if K==0:
        if N*M>0:
            return (-1)
        else:
            return ("0"*N+"1"*M)
    Ncash=N
    Mcash=M
    A=[0 for i in range(K+1)]
    for i in range(1,K+1,2):
        A[i]+=1
        M-=1
    if M<0:
        return -1
    A[K-(K+1)%2]+=M
    for i in range(0,K+1,2):
        A[i]+=1
        N-=1
    if N<0:
        tmp=solve(Ncash,Mcash-1,K-1)
        if tmp==-1:
            return tmp
        else:
            return "1"+tmp
    A[0]+=N
    res=[]
    for i in range(K+1):
        res.append(str(i%2)*A[i])
    return "".join(res)

n,m,k=map(int,input().split())
print(solve(n,m,k))