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())

s =''
       
for i in range(x-1):
    s+=str(int(i%2==1))
    if i%2==1:
        b-=1
    else:
        a-=1
if len(s)!=0:
    if s[-1] == '1':
        s+='0'*a
        s+='1'*b
    elif s[-1]=='0':
        s+='1'*b
        s+='0'*a
elif len(s)==0:
    s+='0'*a
    s+='1'*b
    
count = 0
for i in range(len(s)-1):
    if s[i]!=s[i+1]:
        count+=1
if count+1==x:
    o = s[-1]
    s = o+s[:len(s)-1]
print(s)