You are given three integers a, b and x. Your task is to construct a binary string s of length n = a + b such that there are exactly a zeroes, exactly b ones and exactly x indices i (where 1 ≤ i < n) such that s_i ≠ s_{i + 1}. It is guaranteed that the answer always exists.

For example, for the string "01010" there are four indices i such that 1 ≤ i < n and s_i ≠ s_{i + 1} (i = 1, 2, 3, 4). For the string "111001" there are two such indices i (i = 3, 5).

Recall that binary string is a non-empty sequence of characters where each character is either 0 or 1.

Input

The first line of the input contains three integers a, b and x (1 ≤ a, b ≤ 100, 1 ≤ x < a + b).

Output

Print only one string s, where s is any binary string satisfying conditions described above. It is guaranteed that the answer always exists.

Examples

Input

2 2 1


Output

1100


Input

3 3 3


Output

101100


Input

5 3 6


Output

01010100

Note

All possible answers for the first example: 

  * 1100; 
  * 0011. 



All possible answers for the second example: 

  * 110100; 
  * 101100; 
  * 110010; 
  * 100110; 
  * 011001; 
  * 001101; 
  * 010011; 
  * 001011. 
a,b,x=map(int,(input().split(" ")))
N=a+b
N1=x
if(a>b):
    S=['0']
else:
    S=['1']
for i in range(1,N1+1):
    if(S[i-1]=='0'):
        S.append('1')
    else:
        S.append('0')
W1=S.count('0')
W2=S.count('1')
s=[]
s1=[]
if(W1!=a):
    w=a-W1
    s=['0']*w
if(W2!=b):
    w=b-W2
    s1=['1']*w
if(S[0]=='0'):
    S=s+S
    if(S[-1]=='1'):
        S=S+s1
    else:
        if(len(s1)!=0):
            S[-1]='1'
            s1[-1]='0'
            S=S+s1
else:
    S=s1+S
    if(S[-1]=='0'):
        S=S+s
    else:
        if(len(s)!=0):
            S[-1]='0'
            s[-1]='1'
            S=S+s
    
print("".join(S))
    